In 1967, New Zealander Burt Munro set the world record for an Indian motorcycle, on the Bonneville Salt Flats in Utah, of 183.58 mi/h. The one-way course was 5.00 mi long. Acceleration rates are often described by the time it takes to reach 60.0 mi/h from rest. If this time was 4.00 s, and Burt accelerated at this rate until he reached his maximum speed, how long did it take Burt to complete the course?

Solution:

There are two parts to the race: an acceleration part and a constant speed part.

For the acceleration part:

We are given the following values: v_0=0 \ \text{mph} ; v_f=60 \ \text{mph}; and \Delta t=4.00 \ \text{s} .

First, we need to determine how long (both in distance and time) it takes the motorcycle to finish accelerating. During acceleration, the value of the acceleration is given by

a=\frac{60\:\text{mph}}{4\:\text{s}}

To compute for the time it takes to reach its maximum velocity, we are going to use the formula

v_f=v_0+at

Solving for time t in terms of the other variables

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

Substituting the given values to solve for t_1, the time it takes to accelerate from rest to maximum velocity:

\begin{align*}
t_1 & =\frac{v_f-v_0}{a} \\
t_1 & =\frac{183\:\text{mph}-0\:\text{mph}}{\left(\frac{60\:\text{mph}}{4\:\text{s}}\right)} \\
t_1 & =12.2\:\text{s}
\end{align*}

Since we have a constant acceleration, the distance traveled \Delta x_1 during this period is computed using the formula

\begin{align*}
\Delta x_1 & =v_{ave}t \\
\Delta x_1 & =\left(\frac{v_f+v_0}{2}\right)t \\
\end{align*}

Substituting the given values:

\begin{align*}
\Delta x_1 & =\left(\frac{v_f+v_0}{2}\right)t \\
\Delta x_1 & =\left(\frac{183\:\text{mph}+0\:\text{mph}}{2}\right)\left(12.2\:\text{s}\right) \\
\Delta x_1 & =\left(91.5\:\text{mph}\right)\left(\frac{1\:\text{hr}}{3600\:\text{s}}\right)\left(12.2\:\text{s}\right) \\
\Delta x_1 & =0.31\:\text{mi}
\end{align*}

For the constant speed part:

For the next part of the motion, the speed is constant.

We are given the following values: \Delta x_2=5.0\:\text{mi}-0.3\:\text{mi}=4.7\:\text{mi} .

We are going to solve t_2, the time spent on the course at max speed using the formula

\Delta x_2=v_{max}t_2

Solving for t_2 in terms of the other variables:

t_2=\frac{\Delta x_2}{v_{max}}

Substituting the given values:

\begin{align*}
t_2 & =\frac{\Delta x_2}{v_{max}} \\
t_2  & =\frac{4.7\:\text{mi}}{183\:\text{mph}} \\
t_2  & =\left(0.026\:\text{h}\right)\left(\frac{3600\:\text{s}}{1\:\text{h}}\right) \\
t_2  &=92\:\text{s}
\end{align*}

For the whole course:

So, the total time is

\begin{align*}
t_{total}&=t_1+t_2 \\
t_{total}& =12.2\:\text{s}+\:92\:\text{s} \\
t_{total}& =104\:\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

\end{align*}