Question

In 1990, Dave Campos of the United States rode a special motorcycle called the Easyrider at an average speed of 518 km/h. Suppose that at some point Campos steadily decreases his speed from 100.0 percent to 60.0 percent of his average speed during an interval of 2.00 min. What is the distance traveled during that time interval?​

Answer 1

The distance travelled during the given time can be found out by using the equations of motion.

The distance traveled during the time interval is "13810.8 m".

First, we will find the deceleration of the motorcycle by using the first equation of motion:

$v_f=v_it+at$

where,

vi = initial velocity = (518 km/h)$(\frac{1\ h}{3600\ s})(\frac{1000\ m}{1\ km})$ = 143.89 m/s

vf = final veocity = 60 % of 143.89 m/s = (0.6)(143.89 m/s) = 86.33 m/s

a = deceleration = ?

t =time interval = 2 min = 120 s

Therefore,

$86.33\ m/s = 143.89\ m/s + a(120\ s)\\\\a = \frac{86.33\ m/s - 143.89\ m/s}{120\ s}$

a = -0.48 m/s²

Now, we will use the second equation of motion to find out the distance traveled (s):

$s = v_it+\frac{1}{2}at^2\\\\s = (143.89\ m/s)(120\ s)+\frac{1}{2}(-0.48\ m/s^2)(120\ s)^2\\\\s = 17266.8\ m - 3456\ m$

s = 13810.8 m = 13.81 km

Learn more about the equations of motion here:

brainly.com/question/20594939?referrer=searchResults

The attached picture shows the equations of motion.