Question

A police car is traveling at a velocity of 19.0 m/s due north, when a car zooms by at a constant velocity of 42.0 m/s due north. After a reaction time 0.500 s the policeman begins to pursue the speeder with an acceleration of 5.00 m/s^2. Including the reaction time, how long does it take for the police car to catch up with the speeder?

Answer 1

The police car will catch up the speeder in a time when both will cover a same distance.

the distance covered by him in time t will be given by

s= vt = 42 m/s *t

The distance covered by the police car during the reacti

on time

s1 = 19 m/s ( 0.5) = 9.5 m

The distance covered by the police car in the remaining time t-0.500 is given by the

second equation of motion

s2 = 19( t-0.5) + 0.5 ( 5) ( t-0.5)^2

total distance covered by teh police car in will be

s1+ s2 = 9.5 m +19( t-0.5) + 0.5 ( 5) ( t-0.5)^2

=9.5 m + 19 t -9.5 m + 2.5 ( t- 0.5)^2

= 19 t + 2.5 ( t- 0.5) ^2

As after time t the police car catches the speeder hence the distance covered will be the

same we have

s1 + s2 = s

19 t + 2.5 ( t-0.5)^2 = 42 t

(t-0.5)^2 = (42 -19)t/2

t^2 -1 t + 0.25 = 11.5 t

t^2 -12.5 t +0.25 = 0

solving quadratic equation

t = 12.479 s