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 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 (in s)?

It took Burt 98.1 s to complete the course.

Explanation:

The equations for position and velocity for an accelerated object moving in a straight line are as follows:

x = x0 + v0 * t + 1/2 * a * t²

v = v0 + a * t

Where

x = position at time t

x0 = initial position

v0 = initial speed

a = acceleration

t = time

v = speed at time t

First, let´s find how much time Burt accelerated to reach his maximum speed. With the equation for velocity, we can calculate the acceleration, since we know that in 4 s Burt reached a speed of 60 mi/h

v = v0 + a * t    (v0 = 0)

v = a * t

v/t = a

60.0 mi/ h * (1 h / 3600 s) / 4.00 s = a

a = 4.17 x 10⁻³ mi/s²

With this acceleration, we can calculate how much time Burt accelerated:

v = a * t

183.58 mi/h * (1 h / 3600 s) = 4.17 x 10⁻³ mi/s² * t

183.58 mi/h * (1 h / 3600 s) / 4.17 x 10⁻³ mi/s² = t

t = 12.2 s

Now let´s find the position of Burt after these 12.2 s using the equation for the position:

x = x0 + v0 * t +1/2 * a * t²        (x0 and v0 = 0 since Burt starts from rest at the origin of the reference system)

x = 1/2 * a * t²

x = 1/2 * a * t² = 4.17 x 10⁻³ mi/s² * (12.2 s)² = 0.621 mi

Burt travels 0.621 mi with a constant acceleration, for the rest of the course, he travels with constant speed (a = 0). Then, the position will be:

x = x0 + v  * t

The speed will be the maximum speed reached and the initial position will be 0.621 mi since from that point Burt travels at constant speed.

Then, we can obtain the time Burt needed to complete the rest of the course:

(x - x0) / v = t

(5.00 mi - 0.621 mi) / 183.58 mi/h = t

t = 85.9 s

total time to complete the course = 85.9 s + 12.2 s = 98.1 s