PHYS101: Introduction to Mechanics

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.