Motion Equations for Constant Acceleration in One Dimension

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.

Figure 2.26

The final position $x$ is given by the equation

$x=x_{0}+\bar{v} t$

To find $x$, we identify the values of $x_{0}, \bar{v}$, and $t$ from the statement of the problem and substitute them into the equation.

Solution

1. Identify the knowns. $\bar{v}=4.00 \mathrm{~m} / \mathrm{s}, \Delta t=2.00 \mathrm{~min}$, and $x_{0}=0 \mathrm{~m}$

2. Enter the known values into the equation.

$x=x_{0}+\bar{v} t=0+(4.00 \mathrm{~m} / \mathrm{s})(120 \mathrm{~s})=480 \mathrm{~m}$

Discussion

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

Example 2.9 Calculating Final Velocity: An Airplane Slowing Down after Landing

An airplane lands with an initial velocity of $70.0 \mathrm{~m} / \mathrm{s}$ and then decelerates at $1.50 \mathrm{~m} / \mathrm{s}^{2}$ for $40.0 \mathrm{~s}$. 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.

Figure 2.28

Solution

1. Identify the knowns. $v_{0}=70.0 \mathrm{~m} / \mathrm{s}, a=-1.50 \mathrm{~m} / \mathrm{s}^{2}, t=40.0 \mathrm{~s}$.

2. Identify the unknown. In this case, it is final velocity, $v_{\mathrm{f}}$.

3. Determine which equation to use. We can calculate the final velocity using the equation $v=v_{0}+a t$.

4. Plug in the known values and solve.

$v=v_{0}+a t=70.0 \mathrm{~m} / \mathrm{s}+\left(-1.50 \mathrm{~m} / \mathrm{s}^{2}\right)(40.0 \mathrm{~s})=10.0 \mathrm{~m} / \mathrm{s}$

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.

Figure 2.29 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.

Example 2.10 Calculating Displacement of an Accelerating Object: Dragsters

Dragsters can achieve average accelerations of $26.0 \mathrm{~m} / \mathrm{s}^{2}$. Suppose such a dragster accelerates from rest at this rate for 5.56 s. How far does it travel in this time?

Figure 2.31 U.S. Army Top Fuel pilot Tony "The Sarge" Schumacher begins a race with a controlled burnout.

Strategy

Draw a sketch.

Figure 2.32

We are asked to find displacement, which is $x$ if we take $x_{0}$ 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 $x=x_{0}+v_{0} t+\frac{1}{2} a t^{2}$ once we identify $v_{0}, a$, and $t$ from the statement of the problem.

Solution

1. Identify the knowns. Starting from rest means that $v_{0}=0, a$ is given as $26.0 \mathrm{~m} / \mathrm{s}^{2}$ and $t$ is given as $5.56 \mathrm{~s}$.

2. Plug the known values into the equation to solve for the unknown $x$ :

$x=x_{0}+v_{0} t+\frac{1}{2} a t^{2}.$

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

$x=\frac{1}{2} a t^{2}.$

Substituting the identified values of $a$ and $t$ gives

$x=\frac{1}{2}\left(26.0 \mathrm{~m} / \mathrm{s}^{2}\right)(5.56 \mathrm{~s})^{2}.$

yielding

$x=402 \mathrm{~m}.$

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.

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.

Figure 2.33

The equation $v^{2}=v_{0}^{2}+2 a\left(x-x_{0}\right)$ 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 $v_{0}=0$, since the dragster starts from rest. Then we note that $x-x_{0}=402 \mathrm{~m}$ (this was the answer in Example 2.10). Finally, the average acceleration was given to be $a=26.0 \mathrm{~m} / \mathrm{s}^{2}$

2. Plug the knowns into the equation $v^{2}=v_{0}^{2}+2 a\left(x-x_{0}\right)$ and solve for $v$.

$v^{2}=0+2\left(26.0 \mathrm{~m} / \mathrm{s}^{2}\right)(402 \mathrm{~m}).$

Thus

$v^{2}=2.09 \times 10^{4} \mathrm{~m}^{2} / \mathrm{s}^{2}.$

To get $v$, we take the square root:

$v=\sqrt{2.09 \times 10^{4} \mathrm{~m}^{2} / \mathrm{s}^{2}}=145 \mathrm{~m} / \mathrm{s}.$

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.

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 $7.00 \mathrm{~m} / \mathrm{s}^{2}$, whereas on wet concrete it can decelerate at only $5.00 \mathrm{~m} / \mathrm{s}^{2}$. Find the distances necessary to stop a car moving at $30.0 \mathrm{~m} / \mathrm{s}$ (about $\left.110 \mathrm{~km} / \mathrm{h}\right)$ (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 $0.500 \mathrm{~s}$ to get his foot on the brake.

Strategy

Draw a sketch.

Figure 2.34

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. We shall do this explicitly in the next several examples, using tables to set them off.

Solution for (a)

1. Identify the knowns and what we want to solve for. We know that $v_{0}=30.0 \mathrm{~m} / \mathrm{s} ; v=0$ $a=-7.00 \mathrm{~m} / \mathrm{s}^{2}\left(a\right.$ is negative because it is in a direction opposite to velocity). We take $x_{0}$ to be 0. We are looking for displacement $\Delta x$, or $x-x_{0}$

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

$v^{2}=v_{0}^{2}+2 a\left(x-x_{0}\right).$

This equation is best because it includes only one unknown, $x$. We know the values of all the other variables in this equation. (There are other equations that 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).

3. Rearrange the equation to solve for $x$.

$x-x_{0}=\frac{v^{2}-v_{0}^{2}}{2 a}$

4. Enter known values.

$x-0=\frac{0^{2}-(30.0 \mathrm{~m} / \mathrm{s})^{2}}{2\left(-7.00 \mathrm{~m} / \mathrm{s}^{2}\right)}$

Thus,

$x=64.3 \mathrm{~m} \, \text{on dry concrete.}$

Solution for (b)

This part can be solved in exactly the same manner as Part $A$. The only difference is that the deceleration is $-5.00 \mathrm{~m} / \mathrm{s}^{2}$. The result is

$x_{\text {wet }}=90.0 \mathrm{~m} \text{on wet concrete.}$

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 $\bar{v}=30.0 \mathrm{~m} / \mathrm{s} ; t_{\text {reaction }}=0.500 \mathrm{~s}$ $a_{\text {reaction }}=0.$ We take $x_{0-\text { reaction }}$ to be $0.$ We are looking for $x_{\text {reaction }}.$

2. Identify the best equation to use.

$x=x_{0}+\bar{v} t$ works well because the only unknown value is $x$, which is what we want to solve for.

3. Plug in the knowns to solve the equation.

$x=0+(30.0 \mathrm{~m} / \mathrm{s})(0.500 \mathrm{~s})=15.0 \mathrm{~m}$

This means the car travels $15.0 \mathrm{~m}$ while the driver reacts, making the total displacements in the two cases of dry and wet concrete $15.0 \mathrm{~m}$ greater than if he reacted instantly.

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

$x_{\text {braking }}+x_{\text {reaction }}=x_{\text {total }}$

a. $64.3 \mathrm{~m}+15.0 \mathrm{~m}=79.3 \mathrm{~m}$ when dry

b. $90.0 \mathrm{~m}+15.0 \mathrm{~m}=105 \mathrm{~m}$ when wet

Figure 2.35 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 car merges into freeway traffic on a 200-m-long ramp. If its initial velocity is 10.0 m/s and it accelerates at $2.00 \mathrm{~m} / \mathrm{s}^{2}$, how long does it take to travel the 200 m up the ramp? (Such information might be useful to a traffic engineer).

Strategy

Draw a sketch.

Figure 2.36

We are asked to solve for the time $t$. As before, we identify the known quantities in order to choose a convenient physical relationship (that is, an equation with one unknown, $t$).

Solution

1. Identify the knowns and what we want to solve for. We know that $v_{0}=10 \mathrm{~m} / \mathrm{s} ; a=2.00 \mathrm{~m} / \mathrm{s}^{2} ;$ and $x=200 \mathrm{~m}$

2. We need to solve for $t$. Choose the best equation. $x=x_{0}+v_{0} t+\frac{1}{2} a t^{2}$ works best because the only unknown in the equation is the variable $t$ for which we need to solve.

3. We will need to rearrange the equation to solve for $t$. In this case, it will be easier to plug in the knowns first.

$200 \mathrm{~m}=0 \mathrm{~m}+(10.0 \mathrm{~m} / \mathrm{s}) t+\frac{1}{2}\left(2.00 \mathrm{~m} / \mathrm{s}^{2}\right) t^{2}.$

4. Simplify the equation. The units of meters (m) cancel because they are in each term. We can get the units of seconds (s) to cancel by taking $t=t \mathrm{~s}$, where $t$ is the magnitude of time and $\mathrm{s}$ is the unit. Doing so leaves

$200=10 t+t^{2}.$

5. Use the quadratic formula to solve for $t$.

(a) Rearrange the equation to get 0 on one side of the equation.

$t^{2}+10 t-200=0$

This is a quadratic equation of the form

$a t^{2}+b t+c=0,$

where the constants are $a=1.00, b=10.0$, and $c=-200.$

(b) Its solutions are given by the quadratic formula:

$t=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}.$

This yields two solutions for $t$, which are

$t=10.0 \text { and }-20.0$

In this case, then, the time is $t=t$ in seconds, or

$t=10.0 \mathrm{~s} \text { and }-20.0 \mathrm{~s}.$

A negative value for time is unreasonable, since it would mean that the event happened $20 \mathrm{~s}$ before the motion began. We can discard that solution. Thus,

$t=10.0 \mathrm{~s}.$

Discussion

Whenever an equation contains an unknown squared, there will be two solutions. In some problems both solutions are meaningful, but in others, such as the above, only one solution is reasonable. The 10.0 s answer seems reasonable for a typical freeway on-ramp.