A motorcycle officer hidden at an intersection observes a car driven by an obliv

Question

A motorcycle officer hidden at an intersection observes a car driven by an oblivious driver who ignores a stop sign and continues through the intersection at constant speed. The police officer takes off in pursuit 2.13 s after the car has passed the stop sign. She accelerates at 3.7 m/s2 until her speed is 108 km/h, and then continues at this speed until she catches the car. At that instant, the car is 1.2 km from the intersection.

A) How long did it take for the officer to catch up to the car? (in seconds)

B) How fast was the car traveling. (in mi/h)

Explanation / Answer

time taken to reach 108 km/h, i.e 30 m/s t = v /a = 30 / 3.7 = 8.108108108 seconds

Distance covered in this time t of acceleration = v2 / 2a = 302 / 2x3.7 = 121.6216216 m

distance left out, i.e distance travelled in uniform veloicity = 1200 - 121.6216216 = 1078.378378 m

time taken for this second part of journey where the police travels with uniform speed, T= 1078.378378 / 30

T = 35.94594595 seconds

So, total time taken by the police = t + T =  8.108108108 + 35.94594595 = 44.05405405 seconds

time taken by the officer ~ 44.05 seconds

b.) Total time of journey for the car is 2.13 seconds more thsn that of the motor cycle

so time taken by car = 44.05405405 + 2.13 = 46.18405405 seconds

Distance travelled by the car = 1200 m

Speed of car = distance / time

s = 1200 / 46.10405405 = 25.98299401 m/s = 93.53877845 km/ hr = 58.12244675 miles /hr

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