Question

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

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

Step-by-step solution

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: \(Initial \ Speed \ (m/s) = 212.809 \mathrm{mph} \cdot \frac{1609.34 \mathrm{m}}{1 \mathrm{mile}} \cdot \frac{1 \mathrm{hour}}{3600 \mathrm{s}} \approx 95.116 \mathrm{m/s}\)

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: \(t = \frac{v_f - v_i}{a} \Rightarrow t = \frac{0 - 95.116 \mathrm{m/s}}{-8.0 \mathrm{~m} / \mathrm{s}^{2}} \approx 11.89 \mathrm{s}\)

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.