Question

A car traveling at a constant speed of 30 m/s passes a highway patrol police car which is at rest. The police officer accelerates at a constant rate of 3.0 m/s2 and maintains this rate of acceleration until he pulls next to the speeding car. Assume that the police car starts moving at the moment the speeder passes his car. Determine the – a) time required for the police officer to catch the speeder and – b) distance traveled during the chase.

Answer 1

Answer:

a) 20 seconds

b) 600 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

a) Speeder acceleration is 0 as the speed is constant

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=30\times t+\frac{1}{2}\times 0\times t^2\\\Rightarrow s=30t\ m$

Police car

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=0\times t+\frac{1}{2}\times 3\times t^2\\\Rightarrow s=1.5t^2\ m$

The distance they travel will be equal

$30t=1.5t^2\\\Rightarrow t=\frac{30}{1.5}\\\Rightarrow t=20\ seconds$

Time required by the police car to catch the speeder is 20 seconds

b) The distance traveled by the police car will be 1.5×20² = 600 m