problem solving for acceleration

Use the fact that change equals rate times time, and then add that change to our velocity at the end of the previous problem. Algebra will do the rest for us.

Alternate solution. We don't need no stinkin' conversions with this method. The ratio of eighty to sixty is a simple one, namely 4 3 . From our definition of acceleration, it should be apparent that time is directly proportional to change in velocity when acceleration is constant. Thus…

This is not the answer. It is the time elapsed from the moment when the car began to move. The question was about the additional time needed, so we should subtract the time required to go from zero to sixty. Thus…

∆ t  =  8.8 s − 6.6 s  =  2.2 s

The two methods give essentially the same answer.

Quite simple. Let's do it.

Nothing surprising there except the negative sign. When a vector quantity is negative what does it mean? There are several interpretations of this, but I think mine is the best. When a vector has a negative value, it means that it points in a direction opposite that of the positive vectors. In this problem, since the positive vectors are assumed to point forward (What other direction would a normal car drive?) the acceleration must be backward. Thus the complete answer to this problem is that the car's acceleration is…

a  =  7.16 m/s 2 backward

Although it is common to assign deceleration a negative value, negative acceleration does not automatically imply deceleration. When dealing with vector quantities, any direction can be assumed positive…

up, down, right, left, forward, backward, north, south, east, west

and the corresponding opposite direction assumed negative…

down, up, left, right, backward, forward, south, north, west, east.

It won't matter which you chose as long as you are consistent throughout a problem. Don't learn any rules for assigning signs to particular directions and don't let anyone tell you that a certain direction must be positive or must be negative.

practice problem 2

Acceleration is the rate of change of velocity with time. Since velocity is a vector, this definition means acceleration is also a vector. When it comes to vectors, direction matters as much as size. In a simple one-dimensional problem like this one, directions are indicated by algebraic sign. Every quantity that points away from the batter will be positive. Every quantity that points toward him will be negative. Thus, the ball comes in at −40 m/s and goes out at +50 m/s. If we didn't pay attention to this detail, we wouldn't get the right answer.

practice problem 3

Practice problem 4.

Acceleration

Acceleration , the second derivative of displacement , is defined to be the change of velocity per unit time at a certain instance.

A common misconception is that acceleration implies a POSITIVE change of velocity, while it could also mean a NEGATIVE one.

Formula for Acceleration

Useful Formulae

This article is a stub. Help us out by expanding it .

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3.4 Motion with Constant Acceleration

Learning objectives.

By the end of this section, you will be able to:

• Identify which equations of motion are to be used to solve for unknowns.
• Use appropriate equations of motion to solve a two-body pursuit problem.

You might guess that the greater the acceleration of, say, a car moving away from a stop sign, the greater the car’s displacement in a given time. But, we have not developed a specific equation that relates acceleration and displacement. In this section, we look at some convenient equations for kinematic relationships, starting from the definitions of displacement, velocity, and acceleration. We first investigate a single object in motion, called single-body motion. Then we investigate the motion of two objects, called two-body pursuit problems .

First, let us make some simplifications in notation. Taking the initial time to be zero, as if time is measured with a stopwatch, is a great simplification. Since elapsed time is Δ t = t f − t 0 Δ t = t f − t 0 , taking t 0 = 0 t 0 = 0 means that Δ t = t f Δ t = t f , the final time on the stopwatch. When initial time is taken to be zero, we use the subscript 0 to denote initial values of position and velocity. That is, x 0 x 0 is the initial position and v 0 v 0 is the initial velocity . We put no subscripts on the final values. That is, t is the final time , x is the final position , and v is the final velocity . This gives a simpler expression for elapsed time, Δ t = t Δ t = t . It also simplifies the expression for x displacement, which is now Δ x = x − x 0 Δ x = x − x 0 . Also, it simplifies the expression for change in velocity, which is now Δ v = v − v 0 Δ v = v − v 0 . To summarize, using the simplified notation, with the initial time taken to be zero,

where the subscript 0 denotes an initial value and the absence of a subscript denotes a final value in whatever motion is under consideration.

We now make the important assumption that acceleration is constant . This assumption allows us to avoid using calculus to find instantaneous acceleration. Since acceleration is constant, the average and instantaneous accelerations are equal—that is,

Thus, we can use the symbol a for acceleration at all times. Assuming acceleration to be constant does not seriously limit the situations we can study nor does it degrade the accuracy of our treatment. For one thing, acceleration is constant in a great number of situations. Furthermore, in many other situations we can describe motion accurately by assuming a constant acceleration equal to the average acceleration for that motion. Lastly, for motion during which acceleration changes drastically, such as a car accelerating to top speed and then braking to a stop, motion can be considered in separate parts, each of which has its own constant acceleration.

Displacement and Position from Velocity

To get our first two equations, we start with the definition of average velocity:

Substituting the simplified notation for Δ x Δ x and Δ t Δ t yields

Solving for x gives us

where the average velocity is

The equation v – = v 0 + v 2 v – = v 0 + v 2 reflects the fact that when acceleration is constant, v – v – is just the simple average of the initial and final velocities. Figure 3.18 illustrates this concept graphically. In part (a) of the figure, acceleration is constant, with velocity increasing at a constant rate. The average velocity during the 1-h interval from 40 km/h to 80 km/h is 60 km/h:

In part (b), acceleration is not constant. During the 1-h interval, velocity is closer to 80 km/h than 40 km/h. Thus, the average velocity is greater than in part (a).

Solving for Final Velocity from Acceleration and Time

We can derive another useful equation by manipulating the definition of acceleration:

Substituting the simplified notation for Δ v Δ v and Δ t Δ t gives us

Solving for v yields

Example 3.7

Calculating final velocity.

Second, we identify the unknown; in this case, it is final velocity v f v f .

Last, we determine which equation to use. To do this we figure out which kinematic equation gives the unknown in terms of the knowns. We calculate the final velocity using Equation 3.12 , v = v 0 + a t v = v 0 + a t .

Figure 3.19 is a sketch that shows the acceleration and velocity vectors.

Significance

In addition to being useful in problem solving, the equation v = v 0 + a t v = v 0 + a t gives us insight into the relationships among velocity, acceleration, and time. We can see, for example, that

• Final velocity depends on how large the acceleration is and how long it lasts
• If the acceleration is zero, then the final velocity equals the initial velocity ( v = v 0 ), as expected (in other words, velocity is constant)
• If a is negative, then the final velocity is less than the initial velocity

All these observations fit our intuition. Note that it is always useful to examine basic equations in light of our intuition and experience to check that they do indeed describe nature accurately.

Solving for Final Position with Constant Acceleration

We can combine the previous equations to find a third equation that allows us to calculate the final position of an object experiencing constant acceleration. We start with

Adding v 0 v 0 to each side of this equation and dividing by 2 gives

Since v 0 + v 2 = v – v 0 + v 2 = v – for constant acceleration, we have

Now we substitute this expression for v – v – into the equation for displacement, x = x 0 + v – t x = x 0 + v – t , yielding

Example 3.8

Calculating displacement of an accelerating object.

Second, we substitute the known values into the equation to solve for the unknown:

Since the initial position and velocity are both zero, this equation simplifies to

Substituting the identified values of a and t gives

What else can we learn by examining the equation x = x 0 + v 0 t + 1 2 a t 2 ? x = x 0 + v 0 t + 1 2 a t 2 ? We can see the following relationships:

• Displacement depends on the square of the elapsed time when acceleration is not zero. In Example 3.8 , the dragster covers only one-fourth of the total distance in the first half of the elapsed time.
• If acceleration is zero, then initial velocity equals average velocity ( v 0 = v – ) ( v 0 = v – ) , and x = x 0 + v 0 t + 1 2 a t 2 becomes x = x 0 + v 0 t . x = x 0 + v 0 t + 1 2 a t 2 becomes x = x 0 + v 0 t .

Solving for Final Velocity from Distance and Acceleration

A fourth useful equation can be obtained from another algebraic manipulation of previous equations. If we solve v = v 0 + a t v = v 0 + a t for t , we get

Substituting this and v – = v 0 + v 2 v – = v 0 + v 2 into x = x 0 + v – t x = x 0 + v – t , we get

Example 3.9

Second, we substitute the knowns into the equation v 2 = v 0 2 + 2 a ( x − x 0 ) v 2 = v 0 2 + 2 a ( x − x 0 ) and solve for v :

An examination of the equation v 2 = v 0 2 + 2 a ( x − x 0 ) v 2 = v 0 2 + 2 a ( x − x 0 ) can produce additional insights into the general relationships among physical quantities:

• The final velocity depends on how large the acceleration is and the distance over which it acts.
• For a fixed acceleration, a car that is going twice as fast doesn’t simply stop in twice the distance. It takes much farther to stop. (This is why we have reduced speed zones near schools.)

Putting Equations Together

In the following examples, we continue to explore one-dimensional motion, but in situations requiring slightly more algebraic manipulation. The examples also give insight into problem-solving techniques. The note that follows is provided for easy reference to the equations needed. Be aware that these equations are not independent. In many situations we have two unknowns and need two equations from the set to solve for the unknowns. We need as many equations as there are unknowns to solve a given situation.

Summary of Kinematic Equations (constant a )

Before we get into the examples, let’s look at some of the equations more closely to see the behavior of acceleration at extreme values. Rearranging Equation 3.12 , we have

From this we see that, for a finite time, if the difference between the initial and final velocities is small, the acceleration is small, approaching zero in the limit that the initial and final velocities are equal. On the contrary, in the limit t → 0 t → 0 for a finite difference between the initial and final velocities, acceleration becomes infinite.

Similarly, rearranging Equation 3.14 , we can express acceleration in terms of velocities and displacement:

Thus, for a finite difference between the initial and final velocities acceleration becomes infinite in the limit the displacement approaches zero. Acceleration approaches zero in the limit the difference in initial and final velocities approaches zero for a finite displacement.

Example 3.10

How far does a car go.

• First, we need to identify the knowns and what we want to solve for. We know that v 0 = 30.0 m/s, v = 0, and a = −7.00 m/s 2 ( a is negative because it is in a direction opposite to velocity). We take x 0 to be zero. We are looking for displacement Δ x Δ x , or x − x 0 . Second, we identify the equation that will help us solve the problem. The best equation to use is v 2 = v 0 2 + 2 a ( x − x 0 ) . v 2 = v 0 2 + 2 a ( x − x 0 ) . This equation is best because it includes only one unknown, x . We know the values of all the other variables in this equation. (Other equations would allow us to solve for x , but they require us to know the stopping time, t , which we do not know. We could use them, but it would entail additional calculations.) Third, we rearrange the equation to solve for x : x − x 0 = v 2 − v 0 2 2 a x − x 0 = v 2 − v 0 2 2 a and substitute the known values: x − 0 = 0 2 − ( 30.0 m/s ) 2 2 ( −7.00 m/s 2 ) . x − 0 = 0 2 − ( 30.0 m/s ) 2 2 ( −7.00 m/s 2 ) . Thus, x = 64.3 m on dry concrete . x = 64.3 m on dry concrete .
• This part can be solved in exactly the same manner as (a). The only difference is that the acceleration is −5.00 m/s 2 . The result is x wet = 90.0 m on wet concrete. x wet = 90.0 m on wet concrete.
• When the driver reacts, the stopping distance is the same as it is in (a) and (b) for dry and wet concrete. So, to answer this question, we need to calculate how far the car travels during the reaction time, and then add that to the stopping time. It is reasonable to assume the velocity remains constant during the driver’s reaction time. To do this, we, again, identify the knowns and what we want to solve for. We know that v – = 30.0 m/s v – = 30.0 m/s , t reaction = 0.500 s t reaction = 0.500 s , and a reaction = 0 a reaction = 0 . We take x 0-reaction x 0-reaction to be zero. We are looking for x reaction x reaction . Second, as before, we identify the best equation to use. In this case, x = x 0 + v – t x = x 0 + v – t works well because the only unknown value is x , which is what we want to solve for. Third, we substitute the knowns to solve the equation: x = 0 + ( 3 0.0 m/s ) ( 0.500 s ) = 15 .0 m . x = 0 + ( 3 0.0 m/s ) ( 0.500 s ) = 15 .0 m . This means the car travels 15.0 m while the driver reacts, making the total displacements in the two cases of dry and wet concrete 15.0 m greater than if he reacted instantly. Last, we then add the displacement during the reaction time to the displacement when braking ( Figure 3.23 ), x braking + x reaction = x total , x braking + x reaction = x total , and find (a) to be 64.3 m + 15.0 m = 79.3 m when dry and (b) to be 90.0 m + 15.0 m = 105 m when wet.

Example 3.11

Calculating time.

We need to solve for t . The equation x = x 0 + v 0 t + 1 2 a t 2 x = x 0 + v 0 t + 1 2 a t 2 works best because the only unknown in the equation is the variable t , for which we need to solve. From this insight we see that when we input the knowns into the equation, we end up with a quadratic equation.

We need to rearrange the equation to solve for t , then substituting the knowns into the equation:

We then simplify the equation. The units of meters cancel because they are in each term. We can get the units of seconds to cancel by taking t = t s, where t is the magnitude of time and s is the unit. Doing so leaves

We then use the quadratic formula to solve for t ,

which yields two solutions: t = 10.0 and t = −20.0. A negative value for time is unreasonable, since it would mean the event happened 20 s before the motion began. We can discard that solution. Thus,

Check Your Understanding 3.5

A rocket accelerates at a rate of 20 m/s 2 during launch. How long does it take the rocket to reach a velocity of 400 m/s?

Example 3.12

Acceleration of a spaceship.

Then we substitute v 0 v 0 into v = v 0 + a t v = v 0 + a t to solve for the final velocity:

With the basics of kinematics established, we can go on to many other interesting examples and applications. In the process of developing kinematics, we have also glimpsed a general approach to problem solving that produces both correct answers and insights into physical relationships. The next level of complexity in our kinematics problems involves the motion of two interrelated bodies, called two-body pursuit problems .

Two-Body Pursuit Problems

Up until this point we have looked at examples of motion involving a single body. Even for the problem with two cars and the stopping distances on wet and dry roads, we divided this problem into two separate problems to find the answers. In a two-body pursuit problem , the motions of the objects are coupled—meaning, the unknown we seek depends on the motion of both objects. To solve these problems we write the equations of motion for each object and then solve them simultaneously to find the unknown. This is illustrated in Figure 3.25 .

The time and distance required for car 1 to catch car 2 depends on the initial distance car 1 is from car 2 as well as the velocities of both cars and the acceleration of car 1. The kinematic equations describing the motion of both cars must be solved to find these unknowns.

Consider the following example.

Example 3.13

Cheetah catching a gazelle.

• Equation for the gazelle: The gazelle has a constant velocity, which is its average velocity, since it is not accelerating. Therefore, we use Equation 3.10 with x 0 = 0 x 0 = 0 : x = x 0 + v – t = v – t . x = x 0 + v – t = v – t . Equation for the cheetah: The cheetah is accelerating from rest, so we use Equation 3.13 with x 0 = 0 x 0 = 0 and v 0 = 0 v 0 = 0 : x = x 0 + v 0 t + 1 2 a t 2 = 1 2 a t 2 . x = x 0 + v 0 t + 1 2 a t 2 = 1 2 a t 2 . Now we have an equation of motion for each animal with a common parameter, which can be eliminated to find the solution. In this case, we solve for t : x = v – t = 1 2 a t 2 t = 2 v – a . x = v – t = 1 2 a t 2 t = 2 v – a . The gazelle has a constant velocity of 10 m/s, which is its average velocity. The acceleration of the cheetah is 4 m/s 2 . Evaluating t , the time for the cheetah to reach the gazelle, we have t = 2 v – a = 2 ( 10 m/s ) 4 m/s 2 = 5 s . t = 2 v – a = 2 ( 10 m/s ) 4 m/s 2 = 5 s .
• To get the displacement, we use either the equation of motion for the cheetah or the gazelle, since they should both give the same answer. Displacement of the cheetah: x = 1 2 a t 2 = 1 2 ( 4 m/s 2 ) ( 5 ) 2 = 50 m . x = 1 2 a t 2 = 1 2 ( 4 m/s 2 ) ( 5 ) 2 = 50 m . Displacement of the gazelle: x = v – t = 10 m/s ( 5 ) = 50 m . x = v – t = 10 m/s ( 5 ) = 50 m . We see that both displacements are equal, as expected.

Check Your Understanding 3.6

A bicycle has a constant velocity of 10 m/s. A person starts from rest and begins to run to catch up to the bicycle in 30 s when the bicycle is at the same position as the person. What is the acceleration of the person?

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Speed, Velocity, and Acceleration Problems

What is speed.

Definition of Speed : Speed is the rate of change of distance with time . Speed is different from velocity because it’s not in a specified direction. In this article, you will learn how to solve speed, velocity, and acceleration problems.

Speed, S = Distance (d) / Time (t)

The S.I unit for speed is meter per second (m/s) or ms -1

Non-Uniform or Average Speed : This is a non-steady distance covered by an object at a particular period of time. We can also define non-uniform speed as the type of distance that an object covered at an equal interval of time.

Actual speed: This is also known as the instantaneous speed of an object which is the distance covered by an object over a short interval of time.

You may also like to read:

What is Velocity?

Definition of Velocity: Velocity is the rate of displacement with time. Velocity is the speed of an object in a specified direction. The unit of velocity is the same as that of speed which is meter per second (ms -1 ). We use V as a symbol for velocity.

Uniform Velocity

Definition of uniform velocity: The rate of change of displacement is constant no matter how small the time interval may be. Also, uniform velocity is the distance covered by an object in a specified direction in an equal time interval.

What is Acceleration?

We can also write acceleration as

Uniform Acceleration

What is retardation, equations of motion, solved problems of speed, velocity, and acceleration.

Here are solved problems to help you understand how to calculate speed, velocity, and acceleration:

Distance = 15 m/s x 15 s = 225 m

Final velocity, v = speed = 108 km/h = (108 x 1000m) / (60 x 60s) = 108,000/3,600 = 30m/s

Acceleration = change in velocity/ time = (v-u)/t = (30-0)/15 =30/15 = 2ms -2

Time taken = 1 hour = 1 x 60 x 60s =3,600s

Initial displacement of the car = 80km

The time for 80km is 1hr

Average velocity = total displacement/total time taken = 100km/2hrs = 50km/h

and How to Calculate Velocity Ratio of an Inclined Plane

and we can apply the formula for acceleration that says

Note: The acceleration due to the gravity of a body on the surface of the earth is constant (9.8ms -2 ), even though there may be a slight difference due to the mass and altitude of the body.

maximum height = displacement = 50m

a = (0 – 10 2 )/(2 x 50) = -100/100 = -1 ms -2

Note: Retardation is the negative of acceleration, thus it is written in negative form.

Distance, s =?

s = ut + (1/2)at 2

s = 32 + (1/2) x 80 = 32 + 40 = 72m

Final velocity, v = 0

We already know that acceleration, a = change in velocity/time

A body is traveling at a velocity of 20 m/s and has a mass of 10 kg. How much force is required to change its velocity by 10 m/s in 5 seconds?

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How to Calculate Acceleration

Last Updated: August 1, 2024 Fact Checked

This article was co-authored by Sean Alexander, MS . Sean Alexander is an Academic Tutor specializing in teaching mathematics and physics. Sean is the Owner of Alexander Tutoring, an academic tutoring business that provides personalized studying sessions focused on mathematics and physics. With over 15 years of experience, Sean has worked as a physics and math instructor and tutor for Stanford University, San Francisco State University, and Stanbridge Academy. He holds a BS in Physics from the University of California, Santa Barbara and an MS in Theoretical Physics from San Francisco State University. There are 9 references cited in this article, which can be found at the bottom of the page. This article has been fact-checked, ensuring the accuracy of any cited facts and confirming the authority of its sources. This article has been viewed 1,927,150 times.

Calculating Acceleration from a Force

• Newton’s law can be represented by the equation F net = m x a , where F net is the total force acting on the object, m is the object’s mass, and a is the acceleration of the object.
• When using this equation, keep your units in the metric system . Use kilograms (kg) for mass, newtons (N) for force, and meters per second squared (m/s 2 ) for acceleration.

• For this equation, you will want to convert the mass into kilograms. If the mass you have is in grams simply divide that mass by 1000 to convert to kilograms.

• For example: Let’s say you and your big brother are playing tug-of-war. You pull the rope to the left with a force of 5 newtons while your brother pulls the rope in the opposite direction with a force of 7 newtons. The net force on the rope is 2 newtons to the right, in the direction of your brother.
• In order to properly understand the units, know that 1 newton (N) is equal to 1 kilogram X meter/second squared (kg X m/s 2 ). [5] X Research source

• Force is directly proportional to the acceleration, meaning that a greater force will lead to a greater acceleration.
• Mass is inversely proportional to acceleration, meaning that with a greater mass, the acceleration will decrease.

• For example: A 10 Newton force acts uniformly on a mass of 2 kilograms. What is the object’s acceleration?
• a = F/m = 10/2 = 5 m/s 2

Calculating Average Acceleration from Two Velocities

• Acceleration is a vector quantity, meaning it has both a magnitude and a direction. [9] X Research source The magnitude is the total amount of acceleration whereas the direction is the way in which the object is moving. If it is slowing down the acceleration will be negative.

• Because acceleration has a direction, it is important to always subtract the initial velocity from the final velocity. If you reverse them, the direction of your acceleration will be incorrect.
• Unless otherwise stated in the problem, the starting time is usually 0 seconds.

• If the final velocity is less than the initial velocity, acceleration will turn out to be a negative quantity or the rate at which an object slows down.
• Write the equation: a = Δv / Δt = (v f - v i )/(t f - t i )
• Define the variables: v f = 46.1 m/s, v i = 18.5 m/s, t f = 2.47 s, t i = 0 s.
• Solve: a = (46.1 – 18.5)/2.47 = 11.17 meters/second 2 .
• Define the variables: v f = 0 m/s, v i = 22.4 m/s, t f = 2.55 s, t i = 0 s.
• Solve: a = (0 – 22.4)/2.55 = -8.78 meters/second 2 .

Confirming Your Understanding

• Example Problem: A toy boat with mass 10kg is accelerating north at 2 m/s 2 . A wind blowing due west exerts a force of 100 Newtons on the boat. What is the boat's new northward acceleration?
• Solution: Because the force is perpendicular to the direction of motion, it does not have an effect on motion in that direction. The boat continues to accelerate north at 2 m/s 2 .

• Example Problem: April is pulling a 400 kg container right with a force of 150 newtons. Bob stand on the left of the container and pushes with a force of 200 newtons. A wind blowing left exerts a force of 10 newtons. What is the acceleration of the container?
• Solution: This problem uses tricky language to try and catch you. Draw a diagram and you'll see the forces are 150 newtons right, 200 newtons right, and 10 newtons left. If "right" is the positive direction, the net force is 150 + 200 - 10 = 340 newtons. Acceleration = F / m = 340 newtons / 400 kg = 0.85 m/s 2 .

Calculator, Practice Problems, and Answers

Expert Q&A

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• ↑ Sean Alexander, MS. Academic Tutor. Expert Interview. 14 May 2020.
• ↑ https://www1.grc.nasa.gov/beginners-guide-to-aeronautics/newtons-laws-of-motion/
• ↑ https://www.livescience.com/46560-newton-second-law.html
• ↑ http://www.physicsclassroom.com/Class/newtlaws/u2l2d.cfm
• ↑ http://study.com/academy/lesson/what-is-a-newton-units-lesson-quiz.html
• ↑ https://byjus.com/physics/newtons-second-law-of-motion-and-momentum/
• ↑ http://physics.info/acceleration/
• ↑ http://www.physicsclassroom.com/Class/1DKin/U1L1e.cfm#vttable
• ↑ https://pressbooks.online.ucf.edu/osuniversityphysics/chapter/3-3-average-and-instantaneous-acceleration/

About This Article

To calculate acceleration, use the equation a = Δv / Δt, where Δv is the change in velocity, and Δt is how long it took for that change to occur. To calculate Δv, use the equation Δv = vf - vi, where vf is final velocity and vi is initial velocity. To caltulate Δt, use the equation Δt = tf - ti, where tf is the ending time and ti is the starting time. Once you've calculated Δv and Δt, plug them into the equation a = Δv / Δt to get the acceleration. To learn how to calculate acceleration from a force, read the article! Did this summary help you? Yes No

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Acceleration Formula

What is Acceleration?

When a stationary car starts suddenly, we get pushed up backward, and when brakes are applied, we get pushed forward against our seat, or when our car takes a sharp right turn, we get pushed towards the left. We experience these situations because our car is accelerating.

Simply when there is a change in Velocity, there will be Acceleration. Let’s understand the concept of Acceleration with illustrative examples.

Let’s suppose I have a car moving with a constant Velocity of 90 kmph along a straight line. I can see a helicopter flying at roughly a speed of 20,000 kmph. If I were to ask you that in these two cases, where do you find the Acceleration? Your answer will be surely no because both are moving at a constant pace, so no Acceleration in both cases. 

Now, if I ask you that Acceleration is equal to high speed. What will be your answer? You may say yes, but that’s not true for sure.  Want to know why?  It’s because Acceleration is the rate of change of Velocity. Now, let’s understand the Acceleration formula.

General Formula of Acceleration

We already know that Velocity is a speed with direction; therefore, it is a vector quantity. The Acceleration ‘a’ is given as:

      \[ a  = \frac{\text{Change in Velocity}}{\text{Time Taken}}\]

This formula states that the rate of change in Velocity is the Acceleration, or if the Velocity of an object changes from its initial value ‘u’ to the final value ‘v’, then the expression can be simply written as:

                         \[a =  \frac{(v - u)}{t}\]        

Acceleration Formula in Physics 

In Physics , Acceleration is described as the rate of change of Velocity of an object, irrespective of whether it speeds up or slows down. If it speeds up, Acceleration is taken as positive and if it slows down, the Acceleration is negative. It is caused by the net unbalanced force acting on the object, as per Newton’s Second Law. Acceleration is a vector quantity as it describes the time rate of change of Velocity, which is a vector quantity. Acceleration is denoted by a. Its SI unit is \[\frac{m}{s^{2}}\] and dimensions are \[[M^{0}L^{1}T^{–2}]\].

If \[v_{0}, v_{t}\] and t represents the initial Velocity, final Velocity and the time taken for the change in Velocity, then, the Acceleration is given by:

\[\overrightarrow{a} = \frac{\overrightarrow{v_t} - \overrightarrow{v_0}}{t}\]

In one dimensional motion, we can use;

\[a = \frac{v_t - v_0}{t}\]

If \[\overrightarrow{r} \]represents displacement vector and \[\overrightarrow{v} = \frac{\overrightarrow{\text{d}r}}{\text{d}t}\] represents the velocity, then;

Acceleration:  \[\overrightarrow{a} = \frac{\overrightarrow{\text{d}v}}{\text{d}t} = \frac{\overrightarrow{\text{d}^{2}v}}{\text{d}^{2}t}\]

In one dimensional motion, where x is the displacement, and \[v = \frac{\text{d}r}{\text{d}t}\] is the Velocity, then;

\[a = \frac{\text{d}{v}}{\text{d}t} = \frac{\text{d}^{2}x}{\text{d}^{2}t}\]

A car starts from rest and achieves a speed of 54 \[\frac{km}{h}\] in 3 seconds. Find its Acceleration?

Solution:  \[v_0\] = 0, \[v_t\] = 54 \[\frac{km}{h}\] = 15 \[\frac{m}{s}\], t = 3s, a = ?

Acceleration:  

\[a = \frac{v_t - v_0}{t} = \frac{15 - 0}{3} = 5 \frac{m}{s^{2}}\] 

A body moves along the x- axis according to the relation \[x = 1 – 2 t + 3t^{2}\], where x is in meters and t is in seconds. Find the Acceleration of the body when t = 3 s

We have: \[x = 1 – 2 t + 3t^{2}\]

then; Velocity \[v = \frac{\text{d}x}{\text{d}t} = -2 + 6t \]

Acceleration:  \[v = \frac{\text{d}v}{\text{d}t} =  6  \frac{m}{s^{2}}\].

(We see that the Acceleration is constant here. Therefore, at t = 3s also, its value is 6  \[\frac{m}{s^{2}}\]).

Solved Questions Using Acceleration Formula:

1. What will be the Acceleration of a Car if it Slows from 90 \[\frac{km}{h}\] to a Stop in 10 sec? 

Here, u = 90 \[\frac{km}{h}\] = \[ \frac{90 \times 5}{18} = 25 \frac{m}{s^{2}} \] because initially it was moving at a speed of 90 kmph then reached zero.

Final Velocity ‘v’ = 0 kmph, and t  = 10 seconds

Now, applying the formula here:

\[a  = \frac{(0 - 25)}{10}  = (-) 2.5 \frac{m}{s^{2}} \]

2. A Girl Starts her Motion in a Straight Line at a Velocity of 30 \[\frac{m}{s}\], her Velocity is Changing at a Constant Rate. If She Stops after 60 s, What is her Acceleration?

Answer: Here, the initial Velocity of a girl was 30 \[\frac{m}{s}\] and stops, so her final Velocity will become 0 m/s. Now, the deceleration or retardation occurs, which is just the opposite of Acceleration and it can be determined as:

                  \[  a  = \frac{(0 - 30)}{60} = (-) 0.5 \frac{m}{s^{2}} \]

Question 3: A Car Moves in a Circular Track with a Constant Velocity; will it Experience Acceleration?

Answer: Here, the speed is constant; however, the direction is continuously varying, which means the Velocity is also varying. It states that the car will experience Acceleration.

How to prepare for a test on Acceleration using Vedantu

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FAQs on Acceleration Formula

1. Where can I find sums on Acceleration?

You can find ample examples and sums on Acceleration if you choose to read Acceleration Formula with examples and solved problems on Vedantu. This has a detailed description of what Acceleration is, what its examples are, how it's related to speed and direction.  You will find the kind of sums that will come for the tests here so that you can practice them and secure good marks. Do not skip reading any part as that might cause problems with understanding the concepts later on.

2. How do I clear my concepts related to Velocity and speed?

You should understand the very basics of what each is so that you can answer questions when they are asked. Go through Acceleration Formula with examples and solved problems available on this page on Vedantu as this is the ideal guide for those who are looking to clear their concepts. 

Velocity is the speed with direction. The rate of change in Velocity would be termed Acceleration.  Understanding this is key to solving the sums that come from this chapter and topic. You should go through the entire description along with its formulae and examples. Practice a few sums and then assess your understanding.

3. When a car suddenly halts, why do we get pushed forward against our seat?

As a car suddenly stops,  we tend to get pushed forward as our car starts accelerating. As there is a change in the speed with direction (Velocity), Acceleration occurs.  More about this is provided if you go to  Acceleration Formula with examples and solved problem. This can be found on Vedantu and contains all the relevant details that need to be known. It has explained everything in an extremely simplified manner with the help of examples and is an ideal guidebook for those who are looking to clear their concepts related to Acceleration.

4. Is Acceleration the same as very high speed?

You may be partially correct in stating so but not entirely. Acceleration is more essentially the rate of change of Velocity.  You can read from Acceleration Formula with examples and solve problems on Vedantu and then understand better. It has the description of Acceleration along with its sums and a lot of examples. It has a comprehensive description of the topic and its related concepts so that students do not get confused while learning about the topic. Acceleration as a concept is crucial for the higher classes as the Science and Maths study material that they come across then, will be based on these.

5. Can I skip the chapter and the sums on Acceleration when I am studying?

You should not skip anything at any point in time as it could lead to securing lower grades overall. If you have any queries regarding the concepts on Acceleration and its related problems, you can refer to Acceleration Formula with examples and solved problems on Vedantu’s platform. This will clear all your doubts so that you do not feel tempted to skip this chapter just because it's slightly tricky. Once you have completely scanned this page, you will be able to solve the sums that come for your tests and be more confident in tackling questions that come from the chapter.

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Solved Speed, Velocity, and Acceleration Problems

Simple problems on speed, velocity, and acceleration with descriptive answers are presented for the AP Physics 1 exam and college students. In each solution, you can find a brief tutorial. 

Speed and velocity Problems: 

Problem (1): What is the speed of a rocket that travels $8000\,{\rm m}$ in $13\,{\rm s}$?

Solution : Speed is defined in physics  as the total distance divided by the elapsed time,  so the rocket's speed is \[\text{speed}=\frac{8000}{13}=615.38\,{\rm m/s}\]

Problem (2): How long will it take if you travel $400\,{\rm km}$ with an average speed of $100\,{\rm m/s}$?

Solution : Average speed is the ratio of the total distance to the total time. Thus, the elapsed time is \begin{align*} t&=\frac{\text{total distance}}{\text{average speed}}\\ \\ &=\frac{400\times 10^{3}\,{\rm m}}{100\,{\rm m/s}}\\ \\ &=4000\,{\rm s}\end{align*} To convert it to hours, it must be divided by $3600\,{\rm s}$ which gives $t=1.11\,{\rm h}$.

Problem (3): A person walks $100\,{\rm m}$ in $5$ minutes, then $200\,{\rm m}$ in $7$ minutes, and finally $50\,{\rm m}$ in $4$ minutes. Find its average speed. 

Solution : First find its total distance traveled ($D$) by summing all distances in each section, which gets $D=100+200+50=350\,{\rm m}$. Now, by definition of average speed, divide it by the total time elapsed $T=5+7+4=16$ minutes.

But keep in mind that since the distance is in SI units, so the time traveled must also be in SI units, which is $\rm s$. Therefore, we have\begin{align*}\text{average speed}&=\frac{\text{total distance} }{\text{total time} }\\ \\ &=\frac{350\,{\rm m}}{16\times 60\,{\rm s}}\\ \\&=0.36\,{\rm m/s}\end{align*}

Problem (4): A person walks $750\,{\rm m}$ due north, then $250\,{\rm m}$ due east. If the entire walk takes $12$ minutes, find the person's average velocity. 

Solution : Average velocity , $\bar{v}=\frac{\Delta x}{\Delta t}$, is displacement divided by the elapsed time. Displacement is also a vector that obeys the addition vector rules. Thus, in this velocity problem, add each displacement to get the total displacement . 

In the first part, displacement is $\Delta x_1=750\,\hat{j}$ (due north) and in the second part $\Delta x_2=250\,\hat{i}$ (due east). The total displacement vector is $\Delta x=\Delta x_1+\Delta x_2=750\,\hat{i}+250\,\hat{j}$ with magnitude of  \begin{align*}|\Delta x|&=\sqrt{(750)^{2}+(250)^{2}}\\ \\&=790.5\,{\rm m}\end{align*} In addition, the total elapsed time is $t=12\times 60$ seconds. Therefore, the magnitude of the average velocity is \[\bar{v}=\frac{790.5}{12\times 60}=1.09\,{\rm m/s}\]

Problem (5): An object moves along a straight line. First, it travels at a velocity of $12\,{\rm m/s}$ for $5\,{\rm s}$ and then continues in the same direction with $20\,{\rm m/s}$ for $3\,{\rm s}$. What is its average speed?

Solution: Average velocity is displacement divided by elapsed time, i.e., $\bar{v}\equiv \frac{\Delta x_{tot}}{\Delta t_{tot}}$.

Here, the object goes through two stages with two different displacements, so add them to find the total displacement. Thus,\[\bar{v}=\frac{x_1 + x_2}{t_1 +t_2}\] Again, to find the displacement, we use the same equation as the average velocity formula, i.e., $x=vt$. Thus, displacements are obtained as $x_1=v_1\,t_1=12\times 5=60\,{\rm m}$ and $x_2=v_2\,t_2=20\times 3=60\,{\rm m}$. Therefore, we have \begin{align*} \bar{v}&=\frac{x_1+x_2}{t_1+t_2}\\ \\&=\frac{60+60}{5+3}\\ \\&=\boxed{15\,{\rm m/s}}\end{align*}

Problem (6): A plane flies the distance between two cities in $1$ hour and $30$ minutes with a velocity of $900\,{\rm km/h}$. Another plane covers that distance at $600\,{\rm km/h}$. What is the flight time of the second plane?

Solution: first find the distance between two cities using the average velocity formula $\bar{v}=\frac{\Delta x}{\Delta t}$ as below \begin{align*} x&=vt\\&=900\times 1.5\\&=1350\,{\rm km}\end{align*} where we wrote one hour and a half minutes as $1.5\,\rm h$. Now use again the same kinematic equation above to find the time required for another plane \begin{align*} t&=\frac xv\\ \\ &=\frac{1350\,\rm km}{600\,\rm km/h}\\ \\&=2.25\,{\rm h}\end{align*} Thus, the time for the second plane is $2$ hours and $0.25$ of an hour, which converts to minutes as $2$ hours and ($0.25\times 60=15$) minutes.

Problem (7): To reach a park located south of his jogging path, Henry runs along a 15-kilometer route. If he completes the journey in 1.5 hours, determine his speed and velocity.

Solution:  Henry travels his route to the park without changing direction along a straight line. Therefore, the total distance traveled in one direction equals the displacement, i.e, \[\text{distance traveled}=\Delta x=15\,\rm km\]Velocity is displacement divided by the time of travel \begin{align*} \text{velocity}&=\frac{\text{displacement}}{\text{time of travel}} \\\\ &=\frac{15\,\rm km}{1.5\,\rm h} \\\\ &=\boxed{10\,\rm km/h}\end{align*} and by definition, its average speed is \begin{align*} \text{speed}&=\frac{\text{distance covered}}{\text{time interval}}\\\\&=\frac{15\,\rm km}{1.5\,\rm h}\\\\&=\boxed{10\,\rm km/h}\end{align*} Thus, Henry's velocity is $10\,\rm km/h$ to the south, and its speed is $10\,\rm km/h$. As you can see, speed is simply a positive number, with units but velocity specifies the direction in which the object is moving. 

Problem (8): In 15 seconds, a football player covers the distance from his team's goal line to the opposing team's goal line and back to the midway point of the field having 100-yard-length. Find, (a) his average speed, and (b) the magnitude of the average velocity.

Solution:  The total length of the football field is $100$ yards or in meters, $L=91.44\,\rm m$. Going from one goal's line to the other and back to the midpoint of the field takes $15\,\rm s$ and covers a distance of $D=100+50=150\,\rm yd$. 

Distance divided by the time of travel gets the average speed, \[\text{speed}=\frac{150\times 0.91}{15}=9.1\,\rm m/s\] To find the average velocity, we must find the displacement of the player between the initial and final points. 

The initial point is her own goal line and her final position is the midpoint of the field, so she has displaced a distance of $\Delta x=50\,\rm yd$ or $\Delta x=50\times 0.91=45.5\,\rm m$. Therefore, her velocity is calculated as follows \begin{align*} \text{velocity}&=\frac{\text{displacement}}{\text{time elapsed}} \\\\ &=\frac{45.5\,\rm m}{15\,\rm s} \\\\&=\boxed{3.03\quad \rm m/s}\end{align*} Contrary to the previous problem, here the motion is not in one direction, hence, the displacement is not equal to the distance traveled. Accordingly, the average speed is not equal to the magnitude of the average velocity.

Problem (9): You begin at a pillar and run towards the east (the positive $x$ direction) for $250\,\rm m$ at an average speed of $5\,\rm m/s$. After that, you run towards the west for $300\,\rm m$ at an average speed of $4\,\rm m/s$ until you reach a post. Calculate (a) your average speed from pillar to post, and (b) your average velocity from pillar to post. 

Solution : First, you traveled a distance of $L_1=250\,\rm m$ toward east (or $+x$ direction) at $5\,\rm m/s$. Time of travel in this route is obtained as follows \begin{align*} t_1&=\frac{L_1}{v_1}\\\\ &=\frac{250}{5}\\\\&=50\,\rm s\end{align*} Likewise, traveling a distance of $L_2=300\,\rm m$ at $v_2=4\,\rm m/s$ takes \[t_2=\frac{300}{4}=75\,\rm s\]  (a) Average speed is defined as the distance traveled (or path length) divided by the total time of travel \begin{align*} v&=\frac{\text{path length}}{\text{time of travel}} \\\\ &=\frac{L_1+L_2}{t_1+t_2}\\\\&=\frac{250+300}{50+75} \\\\&=4.4\,\rm m/s\end{align*} Therefore, you travel between these two pillars in $125\,\rm s$ and with an average speed of $4.4\,\rm m/s$. 

(b) Average velocity requires finding the displacement between those two points. In the first case, you move $250\,\rm m$ toward $+x$ direction, i.e., $L_1=+250\,\rm m$. Similarly, on the way back, you move $300\,\rm m$ toward the west ($-x$ direction) or $L_2=-300\,\rm m$. Adding these two gives us the total displacement between the initial point and the final point, \begin{align*} L&=L_1+L_2 \\\\&=(+250)+(-300) \\\\ &=-50\,\rm m\end{align*} The minus sign indicates that you are generally displaced toward the west. 

Finally, the average velocity is obtained as follows: \begin{align*} \text{average velocity}&=\frac{\text{displacement}}{\text{time of travel}} \\\\ &=\frac{-50}{125} \\\\&=-0.4\,\rm m/s\end{align*} A negative average velocity indicating motion to the left along the $x$-axis. 

This speed problem better makes it clear to us the difference between average speed and average speed. Unlike average speed, which is always a positive number, the average velocity in a straight line can be either positive or negative. 

Problem (10): What is the average speed for the round trip of a car moving uphill at 40 km/h and then back downhill at 60 km/h? 

Solution : Assuming the length of the hill to be $L$, the total distance traveled during this round trip is $2L$ since $L_{up}=L_{down}=L$. However, the time taken for going uphill and downhill was not provided. We can write them in terms of the hill's length $L$ as $t=\frac L v$. 

Applying the definition of average speed gives us \begin{align*} v&=\frac{\text{distance traveled}}{\text{total time}} \\\\ &=\frac{L_{up}+L_{down}}{t_{up}+t_{down}} \\\\ &=\cfrac{2L}{\cfrac{L}{v_{up}}+\cfrac{L}{v_{down}}} \end{align*} By reorganizing this expression, we obtain a formula that is useful for solving similar problems in the AP Physics 1 exams. \[\text{average speed}=\frac{2v_{up} \times v_{down}}{v_{up}+v_{down}}\] Substituting the numerical values into this, yields \begin{align*} v&=\frac{2(40\times 60)}{40+60} \\\\ &=\boxed{48\,\rm m/s}\end{align*} What if we were asked for the average velocity instead? During this round trip, the car returns to its original position, and thus its displacement, which defines the average velocity, is zero. Therefore, \[\text{average velocity}=0\,\rm m/s\]

Acceleration Problems

Problem (9): A car moves from rest to a speed of $45\,\rm m/s$ in a time interval of $15\,\rm s$. At what rate does the car accelerate? 

Solution : The car is initially at rest, $v_1=0$, and finally reaches $v_2=45\,\rm m/s$ in a time interval $\Delta t=15\,\rm s$. Average acceleration is the change in velocity, $\Delta v=v_2-v_1$, divided by the elapsed time $\Delta t$, so \[\bar{a}=\frac{45-0}{15}=\boxed{3\,\rm m/s^2} \] 

Problem (10): A car moving at a velocity of $15\,{\rm m/s}$, uniformly slows down. It comes to a complete stop in $10\,{\rm s}$. What is its acceleration?

Solution:  Let the car's uniform velocity be $v_1$ and its final velocity $v_2=0$.   Average acceleration is the difference in velocities divided by the time taken, so we have: \begin{align*}\bar{a}&=\frac{\Delta v}{\Delta t}\\\\&=\frac{v_2-v_1}{\Delta t}\\\\&=\frac{0-15}{10}\\\\ &=\boxed{-1.5\,{\rm m/s^2}}\end{align*}The minus sign indicates the direction of the acceleration vector, which is toward the $-x$ direction.

Problem (11): A car moves from rest to a speed of $72\,{\rm km/h}$ in $4\,{\rm s}$. Find the acceleration of the car.

Solution: Known: $v_1=0$, $v_2=72\,{\rm km/h}$, $\Delta t=4\,{\rm s}$.  Average acceleration is defined as the difference in velocities divided by the time interval between those points \begin{align*}\bar{a}&=\frac{v_2-v_1}{t_2-t_1}\\\\&=\frac{20-0}{4}\\\\&=5\,{\rm m/s^2}\end{align*} In above, we converted $\rm km/h$ to the SI unit of velocity ($\rm m/s$) as \[1\,\frac{km}{h}=\frac {1000\,m}{3600\,s}=\frac{10}{36}\, \rm m/s\] so we get \[72\,\rm km/h=72\times \frac{10}{36}=20\,\rm m/s\] 

Problem (12): A race car accelerates from an initial velocity of $v_i=10\,{\rm m/s}$ to a final velocity of $v_f = 30\,{\rm m/s}$ in a time interval of $2\,{\rm s}$. Determine its average acceleration.

Solution:  A change in the velocity of an object $\Delta v$ over a time interval $\Delta t$ is defined as an average acceleration. Known: $v_i=10\,{\rm m/s}$, $v_f = 30\,{\rm m/s}$, $\Delta t=2\,{\rm s}$. Applying definition of average acceleration, we get \begin{align*}\bar{a}&=\frac{v_f-v_i}{\Delta t}\\&=\frac{30-10}{2}\\&=10\,{\rm m/s^2}\end{align*}

Problem (13): A motorcycle starts its trip along a straight line with a velocity of $10\,{\rm m/s}$ and ends with $20\,{\rm m/s}$ in the opposite direction in a time interval of $2\,{\rm s}$. What is the average acceleration of the car?

Solution:  Known: $v_i=10\,{\rm m/s}$, $v_f=-20\,{\rm m/s}$, $\Delta t=2\,{\rm s}$, $\bar{a}=?$. Using average acceleration definition we have \begin{align*}\bar{a}&=\frac{v_f-v_i}{\Delta t}\\\\&=\frac{(-20)-10}{2}\\\\ &=\boxed{-15\,{\rm m/s^2}}\end{align*}Recall that in the definition above, velocities are vector quantities. The final velocity is in the opposite direction from the initial velocity so a negative must be included.

Problem (14): A ball is thrown vertically up into the air by a boy. After $4$ seconds, it reaches the highest point of its path. How fast does the ball leave the boy's hand?

Solution : At the highest point, the ball has zero speed, $v_2=0$. It takes the ball $4\,\rm s$ to reach that point. In this problem, our unknown is the initial speed of the ball, $v_1=?$. Here, the ball accelerates at a constant rate of $g=-9.8\,\rm m/s^2$ in the presence of gravity.

When the ball is tossed upward, the only external force that acts on it is the gravity force. 

Using the average acceleration formula $\bar{a}=\frac{\Delta v}{\Delta t}$ and substituting the numerical values into this, we will have \begin{gather*} \bar{a}=\frac{\Delta v}{\Delta t} \\\\ -9.8=\frac{0-v_1}{4} \\\\ \Rightarrow \boxed{v_1=39.2\,\rm m/s} \end{gather*} Note that $\Delta v=v_2-v_1$. 

Problem (15): A child drops crumpled paper from a window. The paper hit the ground in $3\,\rm s$. What is the velocity of the crumpled paper just before it strikes the ground? 

Solution : The crumpled paper is initially in the child's hand, so $v_1=0$. Let its speed just before striking be $v_2$. In this case, we have an object accelerating down in the presence of gravitational force at a constant rate of $g=-9.8\,\rm m/s^2$. Using the definition of average acceleration, we can find $v_2$ as below \begin{gather*} \bar{a}=\frac{\Delta v}{\Delta t} \\\\ -9.8=\frac{v_2-0}{3} \\\\ \Rightarrow v_2=3\times (-9.8)=\boxed{-29.4\,\rm m/s} \end{gather*} The negative shows us that the velocity must be downward, as expected!

Problem (16): A car travels along the $x$-axis for $4\,{\rm s}$ at an average velocity of $10\,{\rm m/s}$ and $2\,{\rm s}$ with an average velocity of $30\,{\rm m/s}$ and finally $4\,{\rm s}$ with an average velocity $25\,{\rm m/s}$. What is its average velocity across the whole path?

Solution: There are three different parts with different average velocities. Assume each trip is done in one dimension without changing direction. Thus, displacements associated with each segment are the same as the distance traveled in that direction and is calculated as below: \begin{align*}\Delta x_1&=v_1\,\Delta t_1\\&=10\times 4=40\,{\rm m}\\ \\ \Delta x_2&=v_2\,\Delta t_2\\&=30\times 2=60\,{\rm m}\\ \\ \Delta x_3&=v_3\,\Delta t_3\\&=25\times 4=100\,{\rm m}\end{align*}Now use the definition of average velocity, $\bar{v}=\frac{\Delta x_{tot}}{\Delta t_{tot}}$, to find it over the whole path\begin{align*}\bar{v}&=\frac{\Delta x_{tot}}{\Delta t_{tot}}\\ \\&=\frac{\Delta x_1+\Delta x_2+\Delta x_3}{\Delta t_1+\Delta t_2+\Delta t_3}\\ \\&=\frac{40+60+100}{4+2+4}\\ \\ &=\boxed{20\,{\rm m/s}}\end{align*}

Problem (17): An object moving along a straight-line path. It travels with an average velocity $2\,{\rm m/s}$ for $20\,{\rm s}$ and $12\,{\rm m/s}$ for $t$ seconds. If the total average velocity across the whole path is $10\,{\rm m/s}$, then find the unknown time $t$.

Solution: In this velocity problem, the whole path $\Delta x$ is divided into two parts $\Delta x_1$ and $\Delta x_2$ with different average velocities and times elapsed, so the total average velocity across the whole path is obtained as \begin{align*}\bar{v}&=\frac{\Delta x}{\Delta t}\\\\&=\frac{\Delta x_1+\Delta x_2}{\Delta t_1+\Delta t_2}\\\\&=\frac{\bar{v}_1\,t_1+\bar{v}_2\,t_2}{t_1+t_2}\\\\10&=\frac{2\times 20+12\times t}{20+t}\\\Rightarrow t&=80\,{\rm s}\end{align*}

Note : whenever a moving object, covers distances $x_1,x_2,x_3,\cdots$ in $t_1,t_2,t_3,\cdots$ with constant or average velocities $v_1,v_2,v_3,\cdots$ along a straight-line without changing its direction, then its total average velocity across the whole path is obtained by one of the following formulas

• Distances and times are known:\[\bar{v}=\frac{x_1+x_2+x_3+\cdots}{t_1+t_2+t_3+\cdots}\]
• Velocities and times are known: \[\bar{v}=\frac{v_1\,t_1+v_2\,t_2+v_3\,t_3+\cdots}{t_1+t_2+t_3+\cdots}\]
• Distances and velocities are known:\[\bar{v}=\frac{x_1+x_2+x_3+\cdots}{\frac{x_1}{v_1}+\frac{x_2}{v_2}+\frac{x_3}{v_3}+\cdots}\]

Problem (18): A car travels one-fourth of its path with a constant velocity of $10\,{\rm m/s}$, and the remaining with a constant velocity of $v_2$. If the total average velocity across the whole path is $16\,{\rm m/s}$, then find the $v_2$?

Solution: This is the third case of the preceding note. Let the length of the path be $L$ so \begin{align*}\bar{v}&=\frac{x_1+x_2}{\frac{x_1}{v_1}+\frac{x_2}{v_2}}\\\\16&=\frac{\frac 14\,L+\frac 34\,L}{\frac{\frac 14\,L}{10}+\frac{\frac 34\,L}{v_2}}\\\\\Rightarrow v_2&=20\,{\rm m/s}\end{align*}

Problem (19): An object moves along a straight-line path. It travels for $t_1$ seconds with an average velocity $50\,{\rm m/s}$ and $t_2$ seconds with a constant velocity of $25\,{\rm m/s}$. If the total average velocity across the whole path is $30\,{\rm m/s}$, then find the ratio $\frac{t_2}{t_1}$?

Solution: the velocities and times are known, so we have \begin{align*}\bar{v}&=\frac{v_1\,t_1+v_2\,t_2}{t_1+t_2}\\\\30&=\frac{50\,t_1+25\,t_2}{t_1+t_2}\\\\ \Rightarrow \frac{t_2}{t_1}&=4\end{align*} 

Read more related articles:  

Kinematics Equations: Problems and Solutions

Position vs. Time Graphs

Velocity vs. Time Graphs

In the following section, some sample AP Physics 1 problems on acceleration are provided.

Problem (20): An object moves with constant acceleration along a straight line. If its velocity at instant of $t_1 = 3\,{\rm s}$ is $10\,{\rm m/s}$ and at the moment of $t_2 = 8\,{\rm s}$ is $20\,{\rm m/s}$, then what is its initial speed?

Solution: Let the initial speed at time $t=0$ be $v_0$. Now apply average acceleration definition in the time intervals $[t_0,t_1]$ and $[t_0,t_2]$ and equate them.\begin{align*}\text{average acceleration}\ \bar{a}&=\frac{\Delta v}{\Delta t}\\\\\frac{v_1 - v_0}{t_1-t_0}&=\frac{v_2-v_0}{t_2-t_0}\\\\ \frac{10-v_0}{3-0}&=\frac{20-v_0}{8-0}\\\\ \Rightarrow v_0 &=4\,{\rm m/s}\end{align*} In the above, $v_1$ and $v_2$ are the velocities at moments $t_1$ and $t_2$, respectively. 

Problem (21): For $10\,{\rm s}$, the velocity of a car that travels with a constant acceleration, changes from $10\,{\rm m/s}$ to $30\,{\rm m/s}$. How far does the car travel?

Solution: Known: $\Delta t=10\,{\rm s}$, $v_1=10\,{\rm m/s}$ and $v_2=30\,{\rm m/s}$. 

Method (I) Without computing the acceleration: Recall that in the case of constant acceleration, we have the following kinematic equations for average velocity and displacement:\begin{align*}\text{average velocity}:\,\bar{v}&=\frac{v_1+v_2}{2}\\\text{displacement}:\,\Delta x&=\frac{v_1+v_2}{2}\times \Delta t\\\end{align*}where $v_1$ and $v_2$ are the velocities in a given time interval. Now we have \begin{align*} \Delta x&=\frac{v_1+v_2}{2}\\&=\frac{10+30}{2}\times 10\\&=200\,{\rm m}\end{align*}

Method (II) with computing acceleration: Using the definition of average acceleration, first determine it as below \begin{align*}\bar{a}&=\frac{\Delta v}{\Delta t}\\\\&=\frac{30-10}{10}\\\\&=2\,{\rm m/s^2}\end{align*} Since the velocities at the initial and final points of the problem are given so use the below time-independent kinematic equation to find the required displacement \begin{align*} v_2^{2}-v_1^{2}&=2\,a\Delta x\\\\ (30)^{2}-(10)^{2}&=2(2)\,\Delta x\\\\ \Rightarrow \Delta x&=\boxed{200\,{\rm m}}\end{align*}

Problem (22): A car travels along a straight line with uniform acceleration. If its velocity at the instant of $t_1=2\,{\rm s}$ is $36\,{\rm km/s}$ and at the moment $t_2=6\,{\rm s}$ is $72\,{\rm km/h}$, then find its initial velocity (at $t_0=0$)?

Solution: Use the equality of definition of average acceleration $a=\frac{v_f-v_i}{t_f-t_i}$ in the time intervals $[t_0,t_1]$ and $[t_0,t_2]$ to find the initial velocity as below \begin{align*}\frac{v_2-v_0}{t_2-t_0}&=\frac{v_1-v_0}{t_1-t_0}\\\\ \frac{20-v_0}{6-0}&=\frac{10-v_0}{2-0}\\\\ \Rightarrow v_0&=\boxed{5\,{\rm m/s}}\end{align*}

All these kinematic problems on speed, velocity, and acceleration are easily solved by choosing an appropriate kinematic equation. Keep in mind that these motion problems in one dimension are of the uniform or constant acceleration type. Projectiles are also another type of motion in two dimensions with constant acceleration.

Author:   Dr. Ali Nemati

Date Published: 9/6/2020

Updated: Jun 28,  2023

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Acceleration

Acceleration is how fast velocity changes:

• Speeding up
• Slowing down (also called deceleration )
• Changing direction

It is usually shown as:

m/s 2   "meters per second squared"

What is this " per second squared " thing? An example will help:

A runner accelerates from 5 m/s (5 meters per second) to 6 m/s in just one second

So they accelerate by 1 meter per second per second

See how "per second" is used twice?

It can be thought of as (m/s)/s but is usually written m/s 2

So their acceleration is 1 m/s 2

The formula is:

Acceleration = Change in Velocity (m/s) Time (s)

Example: A bike race!

You are cruising along in a bike race, going a steady 10 meters per second (10 m/s).

Acceleration : Now you start cycling faster! You increase to 14 m/s over the next 2 seconds (still heading in the same direction):

Your speed  increases by 4 m/s , over a time period of 2 seconds , so:

= 4 m/s 2 s = 2 m/s 2

Your speed changes by 2 meters per second per second .

Or more simply "2 meters per second squared".

Example: You are running at 7 m/s, and skid to a halt in 2 seconds.

You went from 7 m/s to 0, so that is a decrease in speed:

= −7 m/s 2 s = −3.5 m/s 2

We don't always say it, but acceleration has direction (making it a vector ):

A car is heading West at 16 m/s . The driver flicks the wheel, and within 4 seconds has the car headed East at 16 m/s .   What is the acceleration?

The numbers are the same, but the direction is different!

Acceleration = From 16 m/s West to 16 m/s East 4 s

From 16 m/s West to 16 m/s East is a total change of 32 m/s towards the East.

Acceleration = 32 m/s East 4 s = 8 m/s 2 East

For more complicated direction changes read vectors .