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dybincka [34]
4 years ago
13

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.
Physics
1 answer:
Lera25 [3.4K]4 years ago
8 0

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

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