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 m/s² 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 traveled at 11.8 m/s until the finish line. How far ahead of him (in meters and in seconds) did the winner finish?

Solution:

We are given the following: v_0=11.5 \ \text{m/s} ; a=0.500 \ \text{m/s}^2; and \Delta t=7.00 \ \text{s}.

Part A

To solve for the final velocity, we are going to use the formula

v_f=v_0+at

Substituting the given values:

\begin{align*}
v_f &=v_0+at\\
v_f&=11.5\ \text{m/s}+\left( 0.500\ \text{m/s}^2 \right)\left( 7.00\ \text{s} \right)\\
v_f&=15.0\ \text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

\end{align*}

Part B

Let  t_{const} be the time it takes to reach the finish line without accelerating:

\begin{align*}
t_{const}&=\frac{x}{v_0}\\
t_{const}&=\frac{300\ \text{m}}{11.5\ \text{m/s}}\\
t_{const}&=26.1\ \text{m/s}
\end{align*}

Now let d be the distance traveled during the 7 seconds of acceleration. We know t=7.00 \ \text{s} so

\begin{align*}
d&=v_0t+\frac{1}{2}at^2\\
d&=\left( 11.5\ \text{m/s} \right)\left( 7.00\ \text{s} \right)+\frac{1}{2}\left( 0.500\ \text{m/s} ^2\right)\left( 7.00\ \text{s} \right)^2\\
d&=92.8\ \text{m}
\end{align*}

Let t' be the time it will take the rider at the constant final velocity to complete the race:

\begin{align*}
t'&=\frac{x-d}{v}\\
t'&=\frac{300\ \text{m}-92.8\ \text{m}}{15.0\ \text{m/s}}\\
t'&=13.8\ \text{s}
\end{align*}

So the total time T it will take the accelerating rider to reach the finish line is 

\begin{align*}
T&=t+t'\\
T&=7.00\ \text{s}+13.8\ \text{s}\\
T&=20.8\ \text{s}
\end{align*}

Finally, let T^{*} be the time saved. So 

\begin{align*}
T^{*}&=26.1\ \text{s}-20.8\ \text{s}\\
T^{*}&={\color{DarkGreen} 5.3\ \text{s}} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

\end{align*}

Part C

For rider 2, we are given the following values: \Delta x'=295 \ \text{m} ; v'=11.8 \ \text{m/s}

Let  t_2 be the time it takes for rider 2 to reach the finish line.

We are going to use the formula

t_2=\frac{\Delta x'}{v'}

Substituting the given values:

\begin{align*}
t_2 & =\frac{x'}{v'} \\
t_2 & =\frac{295\:\text{m}}{11.8\:\text{m/s}} \\
t_2 & =25.0\:\text{s}
\end{align*}

The time difference is

\begin{align*}
\text{time difference} & =t_2-T \\
\text{time difference} & =25.0\:\text{s}-20.817\:\text{s} \\
\text{time difference} & =4.2\:\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

\end{align*}

Therefore, he finishes 4.2 s after the winner.

When the other racer reaches the finish line, he has been traveling at 11.8 m/s for 4.2 seconds, so the other racer finishes

\begin{align*}
\Delta x & =\left(11.8\:\text{m/s}\right)\left(4.2\:\text{s}\right) \\
\Delta x & =49.56\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

\end{align*}

behind the other racer.

Dragsters can actually reach a top speed of 145 m/s in only 4.45 s—considerably less time than given in Example 2.10 and Example 2.11.

(a) Calculate the average acceleration for such a dragster.

(b) Find the final velocity of this dragster starting from rest and accelerating at the rate found in (a) for 402 m (a quarter mile) without using any information on time.

(c) Why is the final velocity greater than that used to find the average acceleration? Hint: Consider whether the assumption of constant acceleration is valid for a dragster. If not, discuss whether the acceleration would be greater at the beginning or end of the run and what effect that would have on the final velocity.

Solution:

We are given the following: v_0=0\ \text{m/s} ; v_f=145 \ \text{m/s}; and \Delta t=4.45 \ \text{sec} .

Part A

To compute for the average acceleration a, we are going to use the formula

a=\frac{\Delta v}{\Delta t}=\frac{v_f-v_0}{\Delta t}

Substituting the given values, we have

\begin{align*}
a & =\frac{v_f-v_0}{\Delta t} \\
a & =\frac{145\:\text{m/s}-0\:\text{m/s}}{4.45\:\text{s}} \\
a & =32.6\:\text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part B

We are given the following: a=32.6 \ \text{m/s}^2 ; v_0=0 \ \text{m/s}; and \Delta x=402 \ \text{m} .

Since we do not have any information on time, we are going to use the formula

\left(v_f\right)^2=\left(v_0\right)^2+2a\Delta x

To compute for the final velocity, we have

v_f=\sqrt{\left(v_0\right)^2+2a\Delta \:x}

Substituting the given values:

\begin{align*}
v_f & =\sqrt{\left(v_0\right)^2+2a\Delta \:x} \\
v_f & =\sqrt{\left(0\:\text{m/s}\right)^2+2\left(32.6\:\text{m/s}^2\right)\left(402\:\text{m}\right)} \\
v_f & =162\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

\end{align*}

Part C

The final velocity is greater than that used to find the average acceleration because the assumption of constant acceleration is not valid for a dragster. A dragster changes gears and would have a greater acceleration in first gear than second gear than third gear, etc. The acceleration would be greatest at the beginning, so it would not be accelerating at 32.6 m/s² during the last few meters, but substantially less, and the final velocity would be less than  162\:\text{m/s}.