A speeder is racing down the road at a constant speed of 90mi/h. A police office

Question

A speeder is racing down the road at a constant speed of 90mi/h. A police officer is having coffee while the speeder zips past her. From the time the speeder passes the officers position, it takes the officer 10 seconds to get into her cruiser and pursue. The officer accelerates at a constant rate of 10/mi/h/s until reaching a top speed of 120mi/h, at which point she travels as a constant speed until she is able to catch the speeder and pull him over.

1) Convert all speeds and accelerations to SI units

2) How Far down the road does the officer catch the speeder

3) How long does it take for the officer to catch up to the speeder

PLEASE ELABORATE,

Thanks, :-)

Explanation / Answer

1.)

1 mile = 1.60934 km = (1609.34) m

Speed of the racer = v1 = 90(1609.34)/3600 = 40.2335 m/s

Acceleration of the officer = 10(1609.34)/3600 = 4.47 m/s/s

Top speed = v2 = 120(1609.34)/3600 = 53.644 m/s

3.)

Distance covered by the speeder in 10 sec = 402.335 m

Let s be the distance down the road

Let the time taken for the officer to catch up to the speeder be t

Thus , using the 2nd Kinemetical Equation

s=ut+(1/2)at^2 = (0.5)(4.47)(t^2) ..............[Equ1]

In time t speeder moves vt=53.664*t m distance

Now s-vt = 402.335

Thus (0.5)(4.47)(t^2) - 53.664t = 402.335

2.235(t^2) - 53.664t - 402.335 = 0

Neglecting the negative value of t,

t= 60 seconds

2.)

By Equ1-

s= (0.5)(4.47)(t^2)

s= (0.5)(4.47)(3600) = 8046 m

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