Question

A motorcycle patrolman is monitoring traffic from behind a billboard along a stretch of road where the speed limit is 96.0 km/hr. He clocks a motorist at 107 km/hr and decides to give chase and award the driver a speeding ticket. By the time he gets onto the highway and up to his chase speed of 131 km/hr, he is 350 m behind the speeder. Determine the amount of time it takes the patrolman to catch the speeder.

Answer 1

Answer:

The time taken is $t = 52.5 \ s$

Explanation:

From the question we are told that

The speed limit is $v__{{l}}} = 96.0 \ km/hr = \frac{96 * 1000}{3600} = 26.7 \ m/s$

The velocity of the motorist is $v_m = 107 \ km/hr = \frac{107 * 1000}{3600} = 29.72 \ m/s$

The chase speed of the motorcycle patrolman is $v = 131 \ km/hr = \frac{131 *1000}{3600} = 36.39 \ m/s$

The relative distance between the motorcycle patrolman and the speeder is d= 350 m

Generally the relative speed between the the motorcycle patrolman and the speeder is mathematically represented as

$v_r = v - v_m$

=> $v_r = 36.39 - 29.72$

=> $v_r = 6.67 \ m/s$

Generally the time taken is mathematically represented as

$t = \frac{v_r}{d}$

=>     $t = \frac{350}{ 6.67}$

=>    $t = 52.5 \ s$