# College Physics 2.39 – Burt Munro setting the world record for an Indian motorcycle

## Solution:

There are two parts to the race: an acceleration part and a constant speed part. 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 $a=\frac{60\:mph}{4\:s}$.

We also know that the maximum velocity is reached by multiplying the acceleration and the time it takes to reach the constant speed. So, we can solve for the time.

$v_{max}=at_1$

$\displaystyle t_1=\frac{v_{max}}{a}$

$\displaystyle t_1=\frac{183\:mph}{\left(\frac{60\:mph\:}{4\:s}\right)}$

$t_1=12.2\:s$

Therefore, we can now solve for the distance traveled for this duration.

$x_1=v_{avg}t_1$

$x_1=\left(\frac{183\:mph+0\:mph}{2}\right)\left(12.2\:s\right)$

$x_1=\left(91.5\:mph\right)\left(\frac{1\:h}{3600\:s}\right)\left(12.2\:s\right)$

$x_1=0.31\:mi$

For the next part of the motion, the velocity is constnat. At this rate, $x_2=5\:mi-0.31\:mi=4.7\:mi$.

Now, we complete the calculation by determining how much time is spent on the course at max speed.

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

$t_2=\frac{4.7\:mi}{183\:mph}$

$t_2=\left(0.026\:h\right)\left(\frac{3600\:s}{1\:h}\right)$

$t_2=92\:s$

So, the total time is

$t_{total}=t_1+t_2$

$t_{total}=12.2\:s+\:92\:s$

$t_{total}=104\:s$