Question

A bicycle racer sprints at the end of a race to clinch a victory. The racer has an initial velocity of 11.5 m/s and accelerates at the rate of 0.650 m/s2 for 7.00 s.What is his final velocity? The racer continues at this velocity to the finish line. If he was 300 m from the finish line when he started to accelerate, how much time did he save? One other racer was 5.00 m ahead when the winner started to accelerate, but he was unable to accelerate and traveled at 11.7 m/s until the finish line. How far ahead of him (in meters and in seconds) did the winner finish?

Answer 1

Question #57994, Physics / Mechanics | Relativity

A bicycle racer sprints at the end of a race to clinch a victory. The racer has an initial velocity of $11.5 \, \text{m/s}$ and accelerates at the rate of $0.650 \, \text{m/s}^2$ for 7.00 s. What is his final velocity? The racer continues at this velocity to the finish line. If he was $300 \, \text{m}$ from the finish line when he started to accelerate, how much time did he save? One other racer was $5.00 \, \text{m}$ ahead when the winner started to accelerate, but he was unable to accelerate and traveled at $11.7 \, \text{m/s}$ until the finish line. How far ahead of him (in meters and in seconds) did the winner finish?

Solution:

a) Final velocity of the racer:

$$\begin{array}{l} v_f = v_0 + a t_{acc}; \\ v_f = 11.5 + 0.650 \times 7 = 16.05 \, \text{m/s} \end{array}$$

b) Distance traveled by racer during acceleration:

$$\begin{array}{l} d_{acc} = v_0 t_{acc} + \frac{a t_{acc}^2}{2}; \\ d_{acc} = 11.5 \times 7 + \frac{0.650 \times 7^2}{2} = 96.43 \, \text{m}; \end{array}$$

Distance traveled by racer with final velocity:

$$\begin{array}{l} d_{fv} = d - d_{acc}; \\ d_{fv} = 300 - 96.43 = 203.57 \, \text{m}; \end{array}$$

Time, spent to travel $d_{fv}$:

$$\begin{array}{l} t_{fv} = \frac{d_{fv}}{v_f}; \\ t_{fv} = \frac{203.57}{16.05} = 12.68 \, \text{s} \end{array}$$

Total time spent to travel $300 \, \text{m}$:

$$\begin{array}{l} t_{total} = t_{acc} + t_{fv} \\ t_{total} = 7 + 12.68 = 19.68 \, \text{s} \end{array}$$

Time required to travel $300 \, \text{m}$ with initial velocity:

$$\begin{array}{l} t_{v0} = \frac{d}{v_0}; \\ t_{v0} = \frac{300}{11.5} = 26.09 \, \text{s}; \end{array}$$

Time saved due to acceleration:

$$\begin{array}{l} \Delta t = t_{v0} - t_{total}; \\ \Delta t = 26.09 - 19.68 = 6.41 \, \text{s} \end{array}$$

c) The second racer was $\Delta d_0 = 5.00 \, \text{m}$ ahead from the first one; therefore, he was 295 m from finish.

Distance traveled by second racer till first racer's finish:

$$\begin{array}{l} d_2 = v_2 t_{total}; \\ d_2 = 11.7 \times 19.68 = 230.26 \, \text{m}; \end{array}$$

Distance behind the leader:

$$\Delta d _ {f} = d - \Delta d _ {0} - d _ {2};$$ $$\Delta d _ {f} = 300 - 5.00 - 230.26 = 64.74 \text{ m}$$

Time behind the winner:

$$\Delta t _ {f} = \frac {\Delta d _ {f}}{v _ {2}} = \frac {64.74}{11.7} = 5.53 \text{ s}$$

Answer: 16.05 m/s; 6.41 s; 64.74 m; 5.53 s.

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