Q2.5-39PE - College Physics (Urone)

Question

In 1967, New Zealander Burt Munro set the world record for an Indian motorcycle, on the Bonneville Salt Flats in Utah, with a maximum speed of\({\bf{183}}.{\bf{58}}{\rm{ }}{\bf{mi}}/{\bf{h}}\). The one-way course was \({\bf{5}}.{\bf{00}}{\rm{ }}{\bf{mi}}\) long. Acceleration rates are often described by the time it takes to reach \({\bf{60}}.{\bf{0}}{\rm{ }}{\bf{mi}}/{\bf{h}}\) from rest. If this time was \({\bf{4}}.{\bf{00}}{\rm{ }}{\bf{s}}\), and Burt accelerated at this rate until he reached his maximum speed, how long did it take Burt to complete the course?

Step-by-Step Solution

Answer

\({\bf{104}}\;{\bf{s}}\).

Acceleration of the runner at t = 4 s

Given data :

• Initial velocity, u = 0.
• Final velocity, v \( = {\bf{60}}.{\bf{0}}\;{\bf{mi}}/{\bf{h}} = {\bf{26}}.{\bf{82}}\;{\bf{m}}/{\bf{s}}\)
• The time is taken to attain the acceleration, T= \({\bf{4}}.{\bf{00}}\;{\bf{s}}\).

The change in velocity gives rise to acceleration. Acceleration can be calculated as:

\(a = \frac{{v - u}}{t}\) 

Initial and final velocities are u and v, and t is the time taken to achieve the final velocity. 

Substituting values in the above expression, we get,

\(\begin{array}{c}a = \frac{{26.82 - 0}}{4}\\ = 6.705\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}\end{array}\)

Hence the acceleration of the body is \(6.705\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}\) 

Time at maximum speed

The maximum speed of the runner is \({\bf{183}}.{\bf{58}}\;{\rm{mi/h}}\).

The maximum velocity in meters/seconds can be calculated as:

\(\begin{array}{c}{V_{\max }} = 183.58\;{\rm{mile/hour}}\\\; = {\rm{ }}82.067\;{\rm{m/s}}\end{array}\) 

The maximum time can be calculated as:

\(\begin{array}{l}{t_{\max }} = \frac{{{v_{\max }} - u}}{a}\\{t_{\max }} = \frac{{82.067 - 0}}{{6.705}}\\{t_{\max }} = 12.239\;{\rm{s}}\end{array}\)

The expression for the distance traveled by him till he reached the maximum speed is,

\(d = ut + \frac{1}{2}a{t^2}\) 

Substituting values in the above expression, we get,

\(\begin{array}{c}d = 0 + \frac{1}{2}(6.705) \times {(12.239)^2}\\ = 502.18\;{\rm{m}}\end{array}\)

Hence the distance calculated is \({\bf{502}}.{\bf{18}}\;{\bf{m}}\).

The remaining distance is,

d = D - s

Here, D is the total length of the one-way course.

Substitute \({\bf{5}}.{\bf{00}}{\rm{ }}{\bf{mi}}\) for D and \({\bf{0}}.{\bf{31}}{\rm{ }}{\bf{mi}}\) for s.

\(\begin{array}{c}d = 5 - 0.31\\ = 4.7\;{\rm{mi}}\\ = {\rm{7563}}{\rm{.92}}\;{\rm{m}}\end{array}\) 

Time can be calculated as:

\(t = \frac{d}{V}\) 

Substituting values in the above expression, we get,

\(\begin{array}{l}t = \frac{{7563.92\;{\rm{m}}}}{{82.067\;{\rm{m/s}}}}\\t = 92.167\;{\rm{s}}\end{array}\)

The time required is \({\bf{92}}.{\bf{167}}{\rm{ }}{\bf{s}}\).

 

The total time \( = 92.167\;{\rm{s}} + 12.239\;{\rm{s }} = 104.4066\;{\rm{s}}\)

Hence, it takes \({\bf{104}}\;{\bf{s}}\) by Burt to complete the course.