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


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. 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