Question

During a test run on an airport runway, a new race car reaches a speed of 258.4 mph from a standing start. The car accelerates with constant acceleration and reaches this speed mark at a distance of \(612.5 \mathrm{~m}\) from where it started. What was its speed after one-fourth, one-half, and three-fourths of this distance?

Short answer

Answer: The car's speeds after one-fourth, one-half, and three-fourths of the distance are approximately 57.716 m/s, 81.719 m/s, and 99.862 m/s, respectively.

Step-by-step solution

Convert the final speed to meters per second (m/s)

To convert the given speed from miles per hour (mph) to meters per second (m/s), we can use the following conversion factors: \(1 \mathrm{~mph} = \frac{1609.34}{3600} \mathrm{~m/s}\). Multiply the given speed (258.4 mph) by this conversion factor: \(258.4 \mathrm{~mph} \cdot \frac{1609.34}{3600} \mathrm{~m/s}\).

Determine the final speed in m/s

By multiplying the given speed by the conversion factor, we find the final speed in meters per second: \(v_f = 258.4 \mathrm{~mph} \cdot \frac{1609.34}{3600} \mathrm{~m/s} \approx 115.492 \mathrm{~m/s}\).

Use the final speed and distance to find the acceleration

We can use the following kinematic equation to find the acceleration given the final speed \(v_f\), initial speed \(v_0\) (0 m/s in this case), distance \(d\), and acceleration \(a\): \(v_f^2 = v_0^2 + 2ad\). Since the car started from a standing position, \(v_0 = 0\). Solve for \(a\) to find the acceleration: \(a = \frac{v_f^2}{2d}\).

Calculate the acceleration

Plug the values for final speed and distance into the acceleration equation: \(a = \frac{(115.492)^2}{2(612.5)} \approx 10.86 \mathrm{~m/s^2}\).

Find the speed at one-fourth, one-half, and three-fourths of the distance

Using the same kinematic equation from step 3, we can find the speeds at the three given distances by replacing \(d\) with the appropriate fractional distance (one-fourth, one-half, and three-fourths of 612.5 meters) and solving for the corresponding final speed value at each distance: 1. One-fourth of distance: \(v_{1/4}^2 = 2a(\frac{1}{4}d)\). Solve for \(v_{1/4}\): \(v_{1/4} = \sqrt{2(10.86)(153.125)} \approx 57.716 \mathrm{~m/s}\). 2. One-half of distance: \(v_{1/2}^2 = 2a(\frac{1}{2}d)\). Solve for \(v_{1/2}\): \(v_{1/2} = \sqrt{2(10.86)(306.25)} \approx 81.719 \mathrm{~m/s}\). 3. Three-fourths of distance: \(v_{3/4}^2 = 2a(\frac{3}{4}d)\). Solve for \(v_{3/4}\): \(v_{3/4} = \sqrt{2(10.86)(459.375)} \approx 99.862 \mathrm{~m/s}\). So, the car's speeds after one-fourth, one-half, and three-fourths of the distance are approximately \(57.716 \mathrm{~m/s}\), \(81.719 \mathrm{~m/s}\), and \(99.862 \mathrm{~m/s}\), respectively.