An unmarked police car traveling a constant 90 km/h is passed by a speeder. Precisely 1.00 s after the speeder passes, the police officer steps on the accelerator. If the police car accelerates uniformly at 3.00 m/s2 and overtakes the speeder after accelerating for 9.00 s , what was the speeder's speed?

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Asked Oct 22, 2019
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An unmarked police car traveling a constant 90 km/h is passed by a speeder. Precisely 1.00 s after the speeder passes, the police officer steps on the accelerator. If the police car accelerates uniformly at 3.00 m/s2 and overtakes the speeder after accelerating for 9.00 s , what was the speeder's speed?

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Step 1

Let up is the initial speed of the police car, us be the speed of the speeder, ap be the acceleration of the police car, t1 be the time after which police car starts accelerating, t2 be the time for which the police car accelerates, s1p, s1s be the distance travelled by police car and the speeder respectively in time t1, and s2p, s2s be the distance travelled by the police car and speeder in time t2.

Write the given values of question.

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90 km/h и, 3(3600) km/h2 = 3 m/s2 1000 1 h 3600 41s 9 h 3600 t9 s

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Step 2

Write the expression to find the distance travelled by police car and speeder in time t1.

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1 h 3600 = (90 km/h = 0.025 km 1 h 3600 =u

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Step 3

Write the expression to find the distance travelled by...

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13(3600) 2 km/h2 = (90 km/h h+ 2 3600 1000 3600 0.3465 km $2 = ,t h 3600

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