Question

Suppose a speeder is traveling with a constant speed of 23 m/s , and that the maximum acceleration of the police car is 3.6 m/s2 .

If a police car is to start from rest and catch the speeder in 15 s or less, what is the maximum head-start distance the speeder can have? Measure time from the moment the police car starts. Express your answer using one significant figure.

Answer 1

v = 23 m/s
a = 3.6 m/s^2
t = 15 s or less

Let the maximum head start distance speeder can have = x
Let police catches in t seconds !!
So,
Distance travelled by Speeder in 15 s,= 23*t
Distance to be travelled by police car, = x + 23*t

Initial Velocity, v = 0
s = u*t + 1/2*a*t^2
x +  23*t = 1/2 * 3.6 * t^2

Now, if we substitute Value of t = 15 then
x +  23*15 = 1/2 * 3.6 * 15^2
x = 60 m

Maximum head start = 60 m

In one sig fig, x = 6.0 * 10^1 m