Questions: A bicycle racer sprints near the end of a race to clinch a victory. The racer has an initial velocity of 10.95 m / s and accelerates at the rate of 0.55 m / s^2 for 5.5 s to accomplish this. What is his final velocity, in meters per second, after this period of acceleration? The racer continues at this velocity to the finish line. If he was 325 m from the finish line when he started to accelerate, how much less time did it take him to finish than if he had not accelerated? One other racer was 6.5 m ahead when the winner started to accelerate, but was too tired to speed up and traveled at 11.6 m / s until the finish line. If the winner continues to cycle at the same speed after crossing the finish line (to celebrate his victory), how far ahead of the loser will the winner be, in meters, when the loser finishes?

Solution Steps

To find the final velocity of the racer after accelerating, we use the formula for final velocity in uniformly accelerated motion:

\[ v_f = v_i + a \cdot t \]

where:

• \( v_i = 10.95 \, \text{m/s} \) is the initial velocity,
• \( a = 0.55 \, \text{m/s}^2 \) is the acceleration,
• \( t = 5.5 \, \text{s} \) is the time of acceleration.

Substituting the values, we get:

\[ v_f = 10.95 + 0.55 \times 5.5 = 10.95 + 3.025 = 13.975 \, \text{m/s} \]

First, calculate the time it would take to finish the race without acceleration. The distance to the finish line is 325 m, and the initial velocity is 10.95 m/s. The time taken without acceleration is:

\[ t_{\text{no accel}} = \frac{325}{10.95} \approx 29.68 \, \text{s} \]

Next, calculate the time taken with acceleration. The distance covered during acceleration is:

\[ d_{\text{accel}} = v_i \cdot t + \frac{1}{2} a \cdot t^2 = 10.95 \times 5.5 + \frac{1}{2} \times 0.55 \times (5.5)^2 \]

\[ d_{\text{accel}} = 60.225 + 8.33625 = 68.56125 \, \text{m} \]

The remaining distance to the finish line after acceleration is:

\[ d_{\text{remaining}} = 325 - 68.56125 = 256.43875 \, \text{m} \]

The time to cover this remaining distance at the final velocity is:

\[ t_{\text{remaining}} = \frac{256.43875}{13.975} \approx 18.34 \, \text{s} \]

The total time with acceleration is:

\[ t_{\text{accel}} = 5.5 + 18.34 = 23.84 \, \text{s} \]

The time saved by accelerating is:

\[ \Delta t = t_{\text{no accel}} - t_{\text{accel}} = 29.68 - 23.84 = 5.84 \, \text{s} \]

The loser travels at a constant speed of 11.6 m/s. The time taken by the loser to finish the race is:

\[ t_{\text{loser}} = \frac{325 + 6.5}{11.6} \approx 28.53 \, \text{s} \]

The winner continues at 13.975 m/s after finishing the race. The time difference between the winner finishing and the loser finishing is:

\[ t_{\text{difference}} = t_{\text{loser}} - t_{\text{accel}} = 28.53 - 23.84 = 4.69 \, \text{s} \]

The distance the winner travels in this time is:

\[ \Delta x = 13.975 \times 4.69 \approx 65.56 \, \text{m} \]

Final Answer