Question

A police car is stopped at a red light facing west. When the light turned green, a sports car speeds past it heading west at with
a constant velocity of 21 m/s. If the police car has a constant acceleration west at 3 m/s, how many seconds will it take to catch
up to the sports car? Answer using only a numeric value, no units.

Answer 1

Answer:

t = 14 seconds

Explanation:

As we know that the speed of the car which passes the police car is 21 m/s

so the distance moved by that car in next "t" seconds is given as

$d = v t$

$d = (21) t$

now we know that police car starts from rest at constant acceleration a = 3 m/s/s

so the distance moved by that car is given as

$d = v_i t + \frac{1}{2} at^2$

$d = 0 + \frac{1}{2}3 t^2$

now if police catch the speeding car after "t" seconds so we have

$21 t = \frac{1}{2}(3) t^2$

$t = 14 s$

Answer 2

Answer:

14 seconds

Explanation:

Distance covered by sports car is measured with 21*t where t is time.

Distance covered by police is measured with at^2/2(t-time, a-acceleration) as it doesnt have starting velocity. If the distances are equal(police catches sport car)

then we have 21t=3/2*t^2 Solving this equation we get t=14 & t=0;

excluding the starting point we have 14 seconds