2.5 Motion Under Constant Acceleration

Rob Pryce

Chapter 2: Describing Movement in One Dimension, 1-D Linear Kinematics

2.5 Motion Under Constant Acceleration

Authors: William Moebs, Samuel Ling, Jeff Sanny
Adapted by: Rob Pryce, Alix Blacklin

Learning Objectives

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

Calculate displacement of an object that is not accelerating, given initial position and velocity.
Calculate final velocity of an accelerating object, given initial velocity, acceleration, and time.
Calculate displacement and final position of an accelerating object, given initial position, initial velocity, time, and acceleration.

Four men racing up a river in their kayaks. — Figure 2.24 Kinematic equations can help us describe and predict the motion of moving objects such as these kayaks racing in Newbury, England. (credit: Barry Skeates, Flickr)

We might know that the greater the acceleration of, say, a car moving away from a stop sign, the greater the displacement in a given time. But we have not developed a specific equation that relates acceleration and displacement. In this section, we develop four convenient equations for kinematic relationships, starting from the definitions of displacement, velocity, and acceleration already covered.

Notation: t, x, v, a

First, let us examine the equations we know so far:

[latex]\bar{v} = \frac {x_f - x_i} {\Delta t}[/latex], and [latex]\bar{a} = \frac {v_f - v_i}{\Delta t}[/latex],

where [latex]\Delta t = t_f - t_i[/latex] and when [latex]t_i = 0[/latex]: [latex]\Delta t = t_f = t[/latex].

Next we 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,

[latex]\bar{a} = a = \text{constant}[/latex]

so we can use the symbol [latex]a[/latex] for acceleration at all times. Assuming acceleration to be constant does not seriously limit the situations we can study nor 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 accurately describe motion by assuming a constant acceleration equal to the average acceleration for that motion. Finally, in motions where acceleration changes drastically, such as an cyclist accelerating to top speed and then slowing down to take a corner, the motion can be considered in separate parts, each of which has its own constant acceleration.

SOLVING FOR DISPLACEMENT ([latex]\Delta x[/latex]) AND FINAL POSITION ([latex]x_f[/latex]) FROM AVERAGE VELOCITY

To get our first equation we can re-arrange the average velocity equation to solve for [latex]x_f[/latex], which gives us:

[latex]x_f = x_i + \bar{v}\Delta t[/latex]

note: when [latex]t_i = 0[/latex], this equation can also be written as [latex]x_f = x_i + \bar{v}t[/latex]

Example 2.8

Calculating Displacement: How far does the jogger run?

A jogger runs down a straight stretch of road with an average velocity of 4.00 m/s for 2.00 min. What is his final position, taking his initial position to be zero?

Strategy

Draw a sketch

Velocity vector arrow labeled v equals 4 point zero zero meters per second over an x axis displaying initial and final positions. Final position is labeled x equals question mark. — Figure 2.25 Reference frame for jogger running at 4.0 m/s for 2:00min.

The final position [latex]x_f[/latex] is given by the equation

[latex]x_f = x_i + \bar{v}\Delta t[/latex]

To find [latex]x_f[/latex], we identify the values of [latex]x_i[/latex], [latex]\bar{v}[/latex], and [latex]\Delta t[/latex] from the statement of the problem and substitute them into the equation.

Solution

Identify the knowns. [latex]\bar{v} = \text {4.00 m/s}[/latex], [latex]\Delta t = \text {2.00 min}[/latex], and [latex]x_i = \text {0.0 m}[/latex].
Notice that velocity is in m/s, however time is in minutes. These must be in the same units (preferably SI), so we 2.00 min to seconds:
[latex]\text {2.00min} \left(\frac {60 s}{1\text{min}}\right) = 120 \text{s}[/latex]
Enter the known values into the equation:
[latex]x_f = x_i + \bar{v}\Delta t = 0 + \left(\text{4.00 m/s}\right)\left(\text{120 s}\right) = 480m[/latex]

Discussion

Velocity and final displacement are both positive, which means they are in the same direction.

The equation [latex]x_f = x_i + \bar{v}\Delta t[/latex] gives insight into the relationship between displacement, average velocity, and time. It shows, for example, that displacement is a linear function of average velocity. (By linear function, we mean that displacement depends on [latex]\bar{v}[/latex] rather than on [latex]\bar{v}[/latex] raised to some other power, such as [latex]\bar{v}^2[/latex]. When graphed, linear functions look like straight lines with a constant slope. In fact, the equation above is simply the equation for a straight line, [latex]y = mx + b[/latex] from introductory algebra). In a running race, for example, someone will go twice as far in a given time if their average velocity is twice that of another runner.

Line graph showing displacement in meters versus average velocity in meters per second. The line is straight with a positive slope. Displacement x increases linearly with increase in average velocity v. — Figure 2.26 There is a linear relationship between displacement and average velocity. For a given time t , an object moving twice as fast as another object will move twice as far as the other object.

SOLVING FOR FINAL VELOCITY ([latex]v_f[/latex]) FROM AVERAGE ACCELERATION

Similarly, we can derive a second equation by re-arranging the average acceleration equation to solve for [latex]v_f[/latex], which gives us:

[latex]v_f = v_i + \bar{a}\Delta t[/latex]

Example 2.9

Calculating Final Velocity: An Airplane Slowing Down after Landing

An airplane lands with an initial velocity of 70.0 m/s and then decelerates at $1 . {50 m/s}^{2}$ for 40.0 s. What is its final velocity?

Strategy

Draw a sketch. We draw the acceleration vector in the direction opposite the velocity vector because the plane is decelerating.

Velocity vector arrow pointing toward the right in the positive x direction. Initial velocity equals seventy meters per second. Final velocity equals question mark. An acceleration vector arrow pointing toward the left labeled a equals negative 1 point 50 meters per second squared. — Figure 2.27 Reference frame for an airplane slowing down

Solution

1. Identify the knowns. [latex]v_i = \text {70.0 m/s}[/latex], [latex]\bar{a} = \text{1.50 m/s}^2[/latex], [latex]\Delta t = \text {40.0 s}[/latex]

2. Identify the unknown. In this case, it is final velocity, [latex]v_f[/latex]_.

3. Determine which equation to use. We can calculate the final velocity using the equation above: [latex]v_f = v_i + \bar{a}\Delta t[/latex].

4. Plug in the known values and solve.

[latex]v_f = v_i + \bar{a}\Delta t = \text {70.0 m/s} + \left(\text{-1.50 m/s}^2\right)\left(\text{40.0s}\right) = \text{10.0 m/s}[/latex]

Discussion

The final velocity is much less than the initial velocity, as desired when slowing down, but still positive. With jet engines, reverse thrust could be maintained long enough to stop the plane and start moving it backward. That would be indicated by a negative final velocity, which is not the case here.

An airplane moving toward the right at two points in time. At time equals 0 the velocity vector arrow points toward the right and is labeled seventy meters per second. The acceleration vector arrow points toward the left and is labeled negative 1 point 5 meters per second squared. At time equals forty seconds, the velocity arrow is shorter, points toward the right, and is labeled ten meters per second. The acceleration vector arrow is still pointing toward the left and is labeled a equals negative 1 point 5 meters per second squared. — Figure 2.28 The airplane lands with an initial velocity of 70.0 m/s and slows to a final velocity of 10.0 m/s before heading for the terminal. Note that the acceleration is negative because its direction is opposite to its velocity, which is positive.

In addition to being useful in problem solving, the equation [latex]v_f = v_i + \bar{a}\Delta t[/latex] gives us insight into the relationships among velocity, acceleration, and time. From it 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 ([latex]v_f = v_i[/latex]) as expected (i.e., velocity is constant)
if [latex]\bar{a}[/latex] is negative, then the final velocity is less than the initial velocity

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

SOLVING FOR FINAL POSITION UNDER CONSTANT ACCELERATION (i.e. when velocity is changing and [latex]\bar{a}\neq 0[/latex])

Our next equation allows us to solve for position given initial velocity and acceleration. The equation is:

[latex]x_f = x_i + v_i\Delta t + \frac{1}{2}\bar{a}\Delta t^2[/latex]

Deriving this equation is a bit more complicated. Here are the steps if you’re interested:

First, we can start with our equation for [latex]v_f[/latex]:

[latex]v_f = v_i + \bar{a}\Delta t[/latex]

Then we can add [latex]v_i[/latex] to each side of the equation and divide by two, which gives:

[latex]\frac{v_f + v_i}{2} = v_i + \frac {1}{2}\bar{a}\Delta t[/latex]

Next, we can assume that when acceleration is constant, [latex]\bar{v}[/latex] is just the simple average of the initial and final velocities:

[latex]\bar{v} = \frac{v_i + v_f}{2}[/latex]

[For example, if you steadily increase your velocity (with constant acceleration) from 3 to 6 m/s, then your average velocity during this steady increase is 4.5 m/s. This can be checked using the above equation: [latex]\frac{3 + 6 \text{m/s}}{2} = \text {4.5 m/s}[/latex] which seems logical.]

This allows us to simply the previous equation to:

[latex]\bar{v} = v_i + \frac{1}{2}\bar{a}\Delta t[/latex]

Finally, we substitute this equation for [latex]\bar{v}[/latex] into the equation for displacement, [latex]x_f = x_i + \bar{v}\Delta t[/latex], yielding:

[latex]x_f = x_i + v_i\Delta t + \frac{1}{2}\bar{a}\Delta t^2[/latex]

Example 2.10

Calculating Displacement of an Accelerating Object: Dragsters

Dragsters can achieve average accelerations of $26 . {0 m/s}^{2}$ . Suppose such a dragster accelerates from rest at this rate for 5.56 s. How far does it travel in this time?

Dragster accelerating down a race track. — Figure 2.30 U.S. Army Top Fuel pilot Tony “The Sarge” Schumacher begins a race with a controlled burnout. (credit: Lt. Col. William Thurmond. Photo Courtesy of U.S. Army.)

Strategy

Draw a sketch.

Acceleration vector arrow pointing toward the right in the positive x direction, labeled a equals twenty-six point 0 meters per second squared. x position graph with initial position at the left end of the graph. The right end of the graph is labeled x equals question mark. — Figure 2.31 Frame of reference for a dragster accelerating

We are asked to find displacement, which is [latex]x_f[/latex]. We can can take [latex]x_i[/latex] to be zero. (Think about it like the starting line of a race. It can be anywhere, but we call it 0 and measure all other positions relative to it.) We can use the equation above, once we identify [latex]v_i[/latex], [latex]\bar{a}[/latex], and [latex]\Delta t[/latex] from the statement of the problem.

Solution

1. Identify the knowns. Starting from rest means that [latex]v_i = 0[/latex], [latex]\bar{a}[/latex] is given as $26.0 {m/s}^{2}$ and [latex]\Delta t[/latex] is given as 5.56 s.

2. Identify the equation with the missing variable ([latex]x_f[/latex]) and plug the known values into the equation to solve for [latex]x_f[/latex]:

[latex]x_f = x_i + v_i\Delta t + \frac{1}{2}\bar{a}\Delta t^2[/latex]

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

[latex]x_f = \frac {1}{2}\bar{a}\Delta t^2[/latex]

Substituting the identified values [latex]\bar{a}[/latex] and [latex]\Delta t[/latex] gives:

[latex]x_f = \frac {1}{2}\left(\text{26.0m/s}^2\right)\left(\text {5.56 x}\right)^2 = \text {402 m}[/latex]

Discussion

If we convert 402 m to miles, we find that the distance covered is very close to one quarter of a mile, the standard distance for drag racing. So the answer is reasonable. This is an impressive displacement in only 5.56 s, but top-notch dragsters can do a quarter mile in even less time than this. Later in the text we can learn how much force the drivers are exposed to during the race.

What else can we learn by examining the equation [latex]x_f = x_i + v_it + \frac{1}{2}\bar{a}\Delta t^2[/latex]? We see that:

displacement depends on the square of the elapsed time when acceleration is not zero. In Example 2.10, the dragster covers only one fourth of the total distance in the first half of the elapsed time.
if acceleration is zero, then the initial velocity equals final velocity ([latex]v_i = v_f[/latex]) and the equation becomes [latex]x_f = x_i + v_i\Delta t[/latex]

SOLVING FOR FINAL VELOCITY UNDER CONSTANT ACCELERATION (i.e. when velocity is changing and [latex]\bar{a}\neq 0[/latex])

Our fourth and last equation can be obtained from another algebraic manipulation of previous equations.

If we solve [latex]v_f = v_i + \bar{a}\Delta t[/latex] for [latex]\Delta t[/latex], we get:

[latex]\Delta t = \frac {v_f - v_i}{\bar{a}}[/latex]

Substituting this and [latex]\bar{v} = \frac {v_i + v_f}{2}[/latex] into [latex]x_f = x_i + \bar{v}\Delta t[/latex], we get our last equation:

[latex]v_f^2 = v_i^2 + 2\bar{a}\Delta x[/latex]

where [latex]\Delta x = x_f - x_i[/latex] $v^{2} = v_{0}^{2} + 2 a (x - x_{0}) (constant a) .$

Example 2.11

Calculating Final Velocity: Dragsters

Calculate the final velocity of the dragster in Example 2.10 without using information about time.

Strategy

Draw a sketch.

Acceleration vector arrow pointing toward the right, labeled twenty-six point zero meters per second squared. Initial velocity equals 0. Final velocity equals question mark. — Figure 2.32

The equation [latex]v_f^2 = v_i^2 + 2\bar{a}\Delta x[/latex] is ideally suited to this task because it relates velocities, acceleration, and displacement, and no time information is required.

Solution

1. Identify the known values. We know that [latex]v_i = 0[/latex], since the dragster starts from rest.

Then we note that [latex]\Delta x = \text {402 m}[/latex] (this was the answer in Example 2.10, and is simply [latex]x_f - x_i[/latex], where [latex]x_f = \text {402m}[/latex] and [latex]x_i = \text {0m}[/latex].

Finally, the average acceleration was given to be [latex]\bar{a} = \text{26.0 m/s}^2[/latex].

2. Plug the knowns into the equation [latex]v_f^2 = v_i^2 + 2\bar{a}\Delta x[/latex] and solve for [latex]v_f[/latex]

[latex]v_f^2 = 0 + 2 \left(26.0 \text{m/s}^2\right)\left(\text {402m}\right)[/latex]

Thus

[latex]v_f^2[/latex] = 20904 m²/s²

To get [latex]v_f[/latex], we take the square root: [latex]v_f = \sqrt{20904 m^2s^2} = 144.6 \text{m/s}[/latex]

Discussion

145 m/s is about 522 km/h or about 324 mi/h, but even this breakneck speed is short of the record for the quarter mile. Also, note that a square root has two values; we took the positive value to indicate a velocity in the same direction as the acceleration.

An examination of the equation [latex]v_f^2 = v_i^2 + 2\bar{a}\Delta x[/latex] can produce further 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 deceleration, a car that is going twice as fast doesn’t simply stop in twice the distance—it takes much further to stop. (This is why we have reduced speed zones near schools.)

Putting Equations Together

In the following examples, we further explore one-dimensional motion, but in situations requiring slightly more algebraic manipulation. The examples also give insight into problem-solving techniques. The box below provides easy reference to the equations needed.

SUMMARY OF KINEMATIC EQUATIONS

[latex]x_f = x_i + \bar{v}\Delta t[/latex]

[latex]v_f = v_i + \bar{a}\Delta t[/latex]

[latex]x_f = x_i + v_it + \frac{1}{2}\bar{a}\Delta t^2[/latex]

[latex]v_f^2 = v_i^2 + 2\bar{a}\Delta x[/latex]

Example 2.12

Calculating Displacement: How Far Does a Car Go When Coming to a Halt?

On dry concrete, a car can decelerate at a rate of $7 . {00 m/s}^{2}$ ², whereas on wet concrete it can decelerate at only $5 . {00 m/s}^{2}$ ². Find the distances necessary to stop a car moving at 30.0 m/s (about 110 km/h) (a) on dry concrete and (b) on wet concrete. (c) Repeat both calculations, finding the displacement from the point where the driver sees a traffic light turn red, taking into account his reaction time of 0.500 s to get his foot on the brake.

Strategy

Draw a sketch.

Initial velocity equals thirty meters per second. Final velocity equals 0. Acceleration dry equals negative 7 point zero zero meters per second squared. Acceleration wet equals negative 5 point zero zero meters per second squared. — Figure 2.33

In order to determine which equations are best to use, we need to list all of the known values and identify exactly what we need to solve for.

Solution for (a)

1. Identify the knowns and what we want to solve for. We know that
[latex]v_i = \text {30.0m/s}[/latex];
[latex]v_f = \text {0 m/s}[/latex];
[latex]\bar{a} = \text {-7.00 m/s}^2[/latex]
([latex]\bar{a}[/latex] is negative because it is in a direction opposite to velocity).
We take [latex]x_i[/latex] to be 0.

We are looking for displacement [latex]\Delta x[/latex], or [latex]x_f - x_i[/latex].

2. Identify the equation that will help up solve the problem. The best equation to use is:

[latex]v_f^2 = v_i^2 + 2\bar{a}\Delta x[/latex]

This equation is best because it includes only one unknown, [latex]x[/latex]. We know the values of all the other variables in this equation. (There are other equations that would allow us to solve for [latex]x[/latex], but they require us to know the stopping time, [latex]\Delta t[/latex], which we do not know. We could use them but it would entail additional calculations.)

3. Rearrange the equation to solve for [latex]x[/latex].

[latex]\Delta x = \frac {v_f^2 - v_i^2}{2\bar{a}}[/latex]

4. Enter known values.

[latex]\Delta x = \frac {0^2 - \left(\text{30m/s}\right)^2}{2\left(\text{-7.0m/s}^2\right)}[/latex]

Thus,

[latex]\Delta x = \text {64.3 m}[/latex] on dry concrete.

Solution for (b)

This part can be solved in exactly the same manner as Part A. The only difference is that the deceleration is $- 5 . {00 m/s}^{2}$ . The result is:

[latex]x_{wet} = \text{90.0m}[/latex] on wet concrete.

Solution for (c)

Once the driver reacts, the stopping distance is the same as it is in Parts 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 that the velocity remains constant during the driver’s reaction time.

1. Identify the knowns and what we want to solve for. We know that
[latex]\bar{v} = \text{30.0m/s}[/latex];
[latex]t_{reaction} = \text {0.500 x}[/latex];
[latex]a_{reaction} = \text 0[/latex].
We take [latex]x_{i-reaction}[/latex] to be 0.

We are looking for [latex]x_{f-reaction}[/latex].

2. Identify the best equation to use.

[latex]x_f = x_i + \bar{v}\Delta t[/latex] works well because the only unknown value is [latex]x_f[/latex], which is what we want to solve for.

3. Plug in the knowns to solve the equation.

[latex]x_f = 0 + \left(\text{30.0 m/s}\right)\left(\text{0.500 x}\right) = \text {15.0 m/s}[/latex]

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.

4. Add the displacement during the reaction time to the displacement when braking.

[latex]x_{braking} + x_{reaction} = x_{total}[/latex]

64.3 m + 15.0 m = 79.3 m when dry
90.0 m + 15.0 m = 105 m when wet

Diagram showing the various braking distances necessary for stopping a car. With no reaction time considered, braking distance is 64 point 3 meters on a dry surface and 90 meters on a wet surface. With reaction time of 0 point 500 seconds, braking distance is 79 point 3 meters on a dry surface and 105 meters on a wet surface. — Figure 2.34 The distance necessary to stop a car varies greatly, depending on road conditions and driver reaction time. Shown here are the braking distances for dry and wet pavement, as calculated in this example, for a car initially traveling at 30.0 m/s. Also shown are the total distances traveled from the point where the driver first sees a light turn red, assuming a 0.500 s reaction time.

Discussion

The displacements found in this example seem reasonable for stopping a fast-moving car. It should take longer to stop a car on wet rather than dry pavement. It is interesting that reaction time adds significantly to the displacements. But more important is the general approach to solving problems. We identify the knowns and the quantities to be determined and then find an appropriate equation. There is often more than one way to solve a problem. The various parts of this example can in fact be solved by other methods, but the solutions presented above are the shortest.

Example 2.13

Calculating Time: A Car Merges into Traffic

Suppose a cyclist is riding down a 200m long hill with an initial velocity is 10.0 m/s. They discover their brakes aren’t working and their bike is accelerating at a constant $2 . {00 m/s}^{2}$ How long will it take them to get to the bottom of the hill?

Strategy

Draw a sketch.

A line segment with ends labeled x subs zero equals zero and x = two hundred. Above the line segment, the equation t equals question mark indicates that time is unknown. Three vectors, all pointing in the direction of x equals 200, represent the other knowns and unknowns. They are labeled v sub zero equals ten point zero meters per second, v equals question mark, and a equals two point zero zero meters per second squared. — Figure 2.35. Reference frame for a cyclist coasting down a hill with an acceleration of 2.0 m/s²

We are asked to solve for the time [latex]\Delta t[/latex]. As before, we identify the known quantities in order to choose an equation with the unknown.

Solution

Identify the knowns and what we want to solve for. We know that
[latex]v_i = \text {10 m/s}[/latex];
[latex]\bar{a} = \text {2.0 m/s}^2[/latex];
[latex]x_i = 0[/latex];
and [latex]x_f = 200m[/latex]
(and [latex]\Delta x = \text {200m}[/latex].
One option would be to use the equation [latex]x_f = x_i + v_it + \frac{1}{2}\bar{a}\Delta t^2[/latex] since it has the unknown quantity, [latex]\Delta t[/latex]. However, as you simplify that equation, you’ll see it has the form of a quadratic equation ([latex]ax^2 + bx + c = 0[/latex]). If you remember it, you can solve this equation using the quadratic formula, however that is not required for this text.
Fortunately, there is another way to solve the problem. These are explained in steps 4 and 5
First, let’s determine the final velocity at the end of the hill using:
[latex]v_f^2 = v_i^2 + 2\bar{a}\Delta x[/latex]
This gives: [latex]v_f^2 = 10^2 + 2\left(2\text {m/s}^2\right)\left(200\text {m}\right)[/latex]
Or [latex]v_f = \sqrt{900} = 30 \text {m/s}[/latex]
Then, we can use the final velocity to solve for time using: [latex]v_f = v_i + \bar{a}\Delta t[/latex]
This gives: [latex]30 \text {m/s} = 10 + \left(2.0\text {m/s}^2\right)\left(\Delta t\right)[/latex]Or [latex]\Delta t = \frac {30 - 10}{2} = 10 \text {s}[/latex]

Discussion

Often there will be multiple ways to solve a problem in kinematics (and biomechanics); sometimes this will involve solving for intermediate variables that can be used in subsequent equations (or using more complex math)

MAKING CONNECTIONS: TAKE-HOME EXPERIMENT—BREAKING NEWS

We have been using SI units of meters per second squared to describe some examples of acceleration or deceleration of cars, runners, and trains. To achieve a better feel for these numbers, one can measure the braking deceleration of a car doing a slow (and safe) stop. Recall that, for average acceleration, [latex]\bar{a} = \frac {\Delta v}{\Delta t}[/latex]. While traveling in a car, slowly apply the brakes as you come up to a stop sign. Have a passenger note the initial speed in miles per hour and the time taken (in seconds) to stop. From this, calculate the deceleration in miles per hour per second. Convert this to meters per second squared and compare with other decelerations mentioned in this chapter. Calculate the distance traveled in braking.

Check Your Understanding

A cyclist can accelerate at a rate of 2.5m/s² during a sprint. Has long does it take them to acceleration from 7.5 m/s to their top speed of 14.5 m/s?

Solution

To answer this, choose an equation that allows you to solve for [latex]\Delta t[/latex] given only [latex]\bar{a}[/latex], [latex]v_i \text{ and, } v_f[/latex].

[latex]v_f = v_i + \bar{a}\Delta t[/latex]

Rearrange to solve for [latex]\Delta t[/latex]

[latex]\Delta t = \frac {v_f - v_i}{\bar{a}} = \frac {14.5\text{m/s} - 7.5 \text{m/s}}{2.5 \text{m/s}^2} = 2.8 \text{s}[/latex]

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Putting Equations Together

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