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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.

Respuesta :

Answer:

The time taken is [tex]t = 52.5 \ s [/tex]

Explanation:

From the question we are told that

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

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

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

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

[tex]v_r = v - v_m[/tex]

=> [tex]v_r = 36.39 - 29.72[/tex]

=> [tex]v_r = 6.67 \ m/s [/tex]

Generally the time taken is mathematically represented as

[tex]t = \frac{v_r}{d}[/tex]

=>     [tex]t =  \frac{350}{ 6.67}[/tex]

=>    [tex]t =  52.5 \  s [/tex]