P01D2017 – Relay Race

Relay Race

The Berkeley relay team is in the lead during the third leg of their race, but the Stanford team is close behind them. In order to win the race, the third runner on the Cal team must pass the baton successfully to the final runner while they are traveling side-by-side at the same velocity. Throughout this problem, the third runner is moving at a constant speed of $v=10\ m/s$. Throughout this entire exam, you may round quantities to one significant figure if you like; for example, you may approximate the magnitude of the acceleration of a projectile as $g=10\ m/s^2$.

1. If the final runner starts from a standing position (i.e. from rest) and then moves at a constant acceleration for $5\ s$, at which point he reaches the speed of the third runner $\left(v=10\ m/s\right)$, what is the magnitude of his acceleration?

We let $a$ denote the acceleration of the final runner whose kinematic equations are given by

\left\{ \begin{array}{c}
\displaystyle{x_f\left(t\right)=\frac{1}{2}at^2+D} \\
\\
v_x\left(t\right)=at \end{array}
\right.

In time $t_1$, the final runner reaches the speed $v=10\ m/s$ of the third runner and therefore his acceleration must satisfy

\begin{aligned}
v_x\left(t_1\right)=v\ \ \ \ &\Rightarrow \ \ \ \ \ at_1=v \\
\\
&\Rightarrow \ \ \ \ \ a=\frac{v}{t_1}
\end{aligned}

The acceleration of the final runner is equal to

\boxed{a=\frac{v}{t_1}=\frac{10}{5}=2\ m/s^2}

2. In that case, what is the average speed of the final runner during the time interval that he accelerates from rest until he reaches his maximum speed of $10\ m/s$?

\boxed{v_{avg}=\frac{v_{ini}+v_{fin}}{2}=\frac{0+10}{2}=5\ m/s}

3. If the final runner starts running at just the right time so that he reaches his final speed running right beside the third runner, then what was the distance $D$ between the two runners right when the final runner started to move?

x_3\left(t\right)=vtx_f\left(t\right)=\frac{1}{2}at^2+D

\begin{aligned}
x_3\left(t_1\right)=x_f\left(t_1\right)\ \ \ \ &\Rightarrow \ \ \ \ \ vt_1=\frac{1}{2}at^2_1+D \\
\\
&\Rightarrow \ \ \ \ \ D=vt_1-\frac{1}{2}at^2_1
\end{aligned}

\boxed{D=vt_1-\frac{1}{2}at^2_1=10\cdot 5-\frac{1}{2}\cdot 2\cdot 5^2=50-25=25\ m}

4. The race is being watched by a commentator riding in a blimp flying over the stadium. If the students in the race are currently running $10\ m/s$ North, and if the blimp is traveling at a velocity of $7\ m/s$ to the North and $4\ m/s$ to the West relative to the ground as shown in the second diagram, then what is the instantaneous velocity of the blimp relative to the runners? Express your answer as a vector.

The blimp is moving with speed $v_{By}=7\ m/s$ to the North while the runners are moving with speed $v=10\ m/s$ to the North. This means that, to the runners, it is as if the blimp were moving at speed $v-v_{By}=3\ m/s$ to the South.

The blimp is moving with speed $v_{Bx}=4\ m/s$ to the West while the runners are not moving to the West whatsoever. This means that, to the runners, the blimp is moving with speed $v_{Bx}=4\ m/s$ to the West.

Thus, the velocity of the blimp relative to the runners is given by

\boxed{{\overrightarrow{v}}_{B/R}=4\hat{x}+3\hat{y}\ \ \ \ \ \left(m/s\right)}

where $\hat{y}$ points to the South as shown in the figure above.

5. If it takes the final runner a total of $10\ s$ to run from the spot where he first grabs the baton all the way around the oval shaped track to that exact location again, then what is his average velocity during that time period? The total length of the track is $100\ m$.