Q2.5-38PE - College Physics (Urone)

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 0.500 for 7.00 s. 

(a) What is his final velocity? 

(b) 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? 

(c) One other racer was 5.00 m ahead when the winner started to accelerate, but he was unable to accelerate, and travelled at 11.8 m/s until the finish line. How far ahead of him (in meters and in seconds) did the winner finish?

Step-by-Step Solution

Answer

a) 15 m/s.

b) 5.274 s.

c) 51.46 m

Given data

• Initial velocity, U = 11.5 m/s.
• Acceleration, a = 0.500 m/s²
• Time required by the racer, T = 7.00 s.
• Distance, d = 300 m

Final velocity of the race

a) 

The final velocity of the racer can be calculated by using the equation of motion as:

V = U + at

Here V is the final velocity, U is the initial velocity, a is the acceleration, and T is the time taken by the racer.

Substituting the values in the above expression, we get,

 $$\begin{gathered} V=11.5+0.5\times 7 \\ V=15 \text{m/s} \end{gathered}$$

Hence the final velocity of the racer is 15 m/s. 

the time saved by the racer

b) 

The distance covered by the racer is 300 m. For this distance, the time taken can be calculated by the equation below:

Here the cyclist starts accelerating from the velocity 15 m/s hence,

Distance traveled by racer during acceleration can be calculated as:

$D_a=ut+\frac{1}{2}a{t}^2$

Substituting the values in the above expression, we get,

$$\begin{gathered} D_a=11.5\times 7+\frac{1}{2}\times (0.5)\times 49 \\ {D}_a=92.75 \text{m} \end{gathered}$$ 

The distance traveled by the racer during acceleration is 92.75 meter

 

Distance traveled by racer with final velocity can be calculated as:

$$\begin{gathered} D_{fv}=300-D_a \\ {D}_{fv}=300-92.75 \\ {D}_{fv}=207.25 \text{m} \end{gathered}$$

The distance traveled by the racer with final velocity is 207.25 m

 

The time spends to travel D_fv distance can be calculated as:

$$\begin{gathered} T_{fv}=\frac{D_{fv}}{V_f} \\ {T}_{fv}=\frac{207.25}{15} \\ {T}_{fv}=13.816 \text{s} \end{gathered}$$

The time taken is 13.816 s. 

The time required to travel 300 m with initial velocity can be calculated as:

$$\begin{gathered} T_0=\frac{D_0}{V_0} \\ {T}_0=\frac{300}{11.5} \\ {T}_0=26.09 \text{s} \end{gathered}$$

Hence The time required to travel 300 meters with an intial velocity is 26.09 S

Total time spent to travel 300 m can be calculated as:

$$\begin{gathered} T_f=T_{fv}+T \\ {T}_f=13.816+7 \\ {T}_f=20.816 \text{s} \end{gathered}$$

The total time spent to travel 300 m is 20.816 s.

Time saved due to acceleration can be calculated as:

$$\begin{gathered} T_s=T_0-T_f \\ {T}_s=26.09-20.816 \\ {T}_s=5.274 \text{s} \end{gathered}$$

Hence the total time saved is 5.274 s.

Distance and time by the second runner

c)

The second racer was Dd = 5.00 m ahead of the first one; therefore, he was 295 m from the finish.

Distance traveled by the second racer till the first racer’s finish:

$$\begin{gathered} D=VT \\ D=11.7\times 20.816 \\ D=243.54 \text{m} \end{gathered}$$

The distance traveled by the second racer is 243.54 m.

Distance behind the winner = 300 m -5 m - 243.54 m = 51.46 m

Hence the winner is 51.46 m ahead of the second person.