<p>11.39 A police officer in a patrol car parked in a 70 km/h speed zone<br />ob

Question

<p>11.39 A police officer in a patrol car parked in a 70 km/h speed zone<br />observes a passing automobile traveling at a slow, constant speed.<br />Believing that the driver of the automobile might be intoxicated,<br />the officer starts his car, accelerates uniformly to 90 km/h in 8 s,<br />and, maintaining a constant velocity of 90 km/h, overtakes the<br />motorist 42 s after the automobile passed him. Knowing that 18 s<br />elapsed before the officer began pursuing the motorist, determine<br />(a) the distance the officer traveled before overtaking the motorist,<br />(b) the motorist&#8217;s speed.</p>

Explanation / Answer

Time for which police is moving = 42-18=24 sec for first 8 s final velocity = 90 km/h =25 m/s so acceleration = {v-u}/t = 25/8 =3.125 m/s^2 distance traveled = ut+at^2/2 = 3.125*64/2 = 100 m Time for which police is moving with constant speed = 24-8 = 16sec distance traveled = 16*25 = 400 m So distance the officer traveled before overtaking the motorist = 400+100 = 500 m Also this is the same distance traveled by motorist in 42 sec so speed of motorist = 500/42 =11.9 m/s = 42.86 km/hr

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