Question

Chagter 02 Problem 8 LİNK TO ft TEXT LINK TO

Answer 1

Let the police car catch up with the other car in a time t secs.

Then the distance traveled by the 2 cars in this time is the same.

The distance traveled by the speeding car in t secs =48t m.

By applying the law of kinematics for motion in one dimension with uniform acceleration we obtain the distance traveled by the police car in a time t'=t-0.8 s =$ut'+1/2at'^{2}=19.8(t-0.8)+1/2*4*(t-0.8)^{2}$

accounting for the reaction time of 0.8s. Here u=19.8 m/s is the initial velocity of the car and a=4 m/s² is its acceleration. The positive direction is taken to be in the direction of motion of the 2 cars i.e. due north.

Now we have $48t=19.8(t-0.8)+1/2*4*(t-0.8)^{2}=2t^{2}-16.6-14.6$ or

$2t^{2}-64.8t-14.6=0$. On solving this quadratic equation we obtain t=39.2 s.