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A police officer at rest at side of highway notices speeder moving at 62 km/h along road.when speeder passes ,officer accelerates 3.0 m/s^2 in pursuit.speeder do not notice until police catch up.

Ahow long will it take for the officer to catch the speeder
Bcalculate the speed of the police officer when he catches the speeder, is this reasonable?
Cnow assume that the police fficer accelerates until the police car is moving 10 km/h faster than the speeder and then moves at a constant velocity until the car catches up to the speeder. how long will it take to catch the speeder?

Verified Answer :

To answer the following questions for this specific problem:

a. 11.48 secs

b. Vp = at3.6 =
311.483.6 = 124.0 km/h

c. 9.1 secs. 

I am hoping that this answer has satisfied your query about
and it will be able to help you.

Answer:

Part a)

\(t = 11.5 s\)

Part b)

\(v_f = 34.44 m/s\)

Part c)

\(T = 23.9 s\)

Explanation:

Part a)

Let after time "t" the officer catches the speeder

So here the distance moved by the speeder and the police must be same

so we have

\(d_{p} = d_s\)

\(d_p = \frac{1}{2}at^2\)

\(d_s = (62 \times \frac{1000}{3600})(t)\)

now we have

\(17.22 t = 1.5 t^2\)

\(t = 11.5 s\)

Part b)

Speed of the police officer when he catches the speeder is given as

\(v_f = v_i + at\)

\(v_f = 0 + 3(11.5)\)

\(v_f = 34.44 m/s\)

Part c)

Police accelerated till his speed become

\(v_f = 62 + 10 = 72 km/h\)

\(v_f = 20 m/s\)

so time taken till it reach this speed

\(v_f = v_i + at\)

\(20 = 0 + 3t\)

\(t = 6.67 s\)

now after this the distance between them

\(d = (17.22\times 6.67) - \frac{1}{2}(3)(6.67)^2\)

\(d = 48 m\)

now this distance will covered in the next few seconds given as

\(d = (v_p - v_s) t\)

\(48 = (20 - 17.22)t\)

\(t = 17.3 s\)

so total time taken by it

\(T = 17.3 + 6.67 = 23.9 s\)

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