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Chapter 2: Problem 85

The fastest speed in NASCAR racing history was \(212.809 \mathrm{mph}\) (reached by Bill Elliott in 1987 at Talladega). If the race car decelerated from that speed at a rate of \(8.0 \mathrm{~m} / \mathrm{s}^{2},\) how far would it travel before coming to a stop?

Short Answer

Expert verified
Answer: The NASCAR race car would travel approximately 565.395 meters before coming to a complete stop.

Step by step solution

01

Convert the initial speed from mph to m/s

To convert the speed from miles per hour to meters per second, we can use the conversion factor: \(1 \mathrm{~mile} = 1609.34 \mathrm{~m}\) and \(1 \mathrm{~hour} = 3600 \mathrm{~s}\). Thus, the conversion factor is \(\frac{1609.34 \mathrm{m}}{1 \mathrm{mile}} \cdot \frac{1 \mathrm{hour}}{3600 \mathrm{s}}\). Applying this to the given speed: \(Initial \ Speed \ (m/s) = 212.809 \mathrm{mph} \cdot \frac{1609.34 \mathrm{m}}{1 \mathrm{mile}} \cdot \frac{1 \mathrm{hour}}{3600 \mathrm{s}} \approx 95.116 \mathrm{m/s}\)
02

Calculate the time taken to come to a stop

We can use the kinematic equation that relates initial speed, final speed, acceleration and time: \(v_f = v_i + at\), where \(v_f\) is the final speed (0 m/s in this case, since the car stops), \(v_i\) is the initial speed, \(a\) is the acceleration (deceleration, which is negative), and \(t\) is the time taken. Rearranging the equation for time, we get: \(t = \frac{v_f - v_i}{a} \Rightarrow t = \frac{0 - 95.116 \mathrm{m/s}}{-8.0 \mathrm{~m} / \mathrm{s}^{2}} \approx 11.89 \mathrm{s}\)
03

Calculate the distance traveled during deceleration

Now that we have the time taken to come to a stop, we can use another kinematic equation to determine the distance traveled during deceleration: \(d = v_i t + \frac{1}{2}a t^2\). Plugging in the values we have: \(d = 95.116 \mathrm{m/s} \cdot 11.89 \mathrm{s} + \frac{1}{2} \cdot (-8.0 \mathrm{~m} / \mathrm{s}^{2}) \cdot (11.89 \mathrm{s})^2 \approx 565.395 \mathrm{m}\) The NASCAR race car would travel approximately \(565.395 \mathrm{m}\) before coming to a complete stop.

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Most popular questions from this chapter

The planet Mercury has a mass that is \(5 \%\) of that of Earth, and its gravitational acceleration is \(g_{\text {mercury }}=3.7 \mathrm{~m} / \mathrm{s}^{2}\) a) How long does it take for a rock that is dropped from a height of \(1.75 \mathrm{~m}\) to hit the ground on Mercury? b) How does this time compare to the time it takes the same rock to reach the ground on Earth, if dropped from the same height? c) From what height would you have to drop the rock on Earth so that the fall- time on both planets is the same?

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The position of an object as a function of time is given as \(x=A t^{3}+B t^{2}+C t+D .\) The constants are \(A=2.1 \mathrm{~m} / \mathrm{s}^{3}\) \(B=1.0 \mathrm{~m} / \mathrm{s}^{2}, C=-4.1 \mathrm{~m} / \mathrm{s},\) and \(D=3 \mathrm{~m}\) a) What is the velocity of the object at \(t=10.0 \mathrm{~s}\) ? b) At what time(s) is the object at rest? c) What is the acceleration of the object at \(t=0.50 \mathrm{~s} ?\) d) Plot the acceleration as a function of time for the time interval from \(t=-10.0 \mathrm{~s}\) to \(t=10.0 \mathrm{~s}\).

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