Question

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. A) how long will it take for the officer to catch the speeder

Answer 1

Use v = v_0 + a*t

Answer 2

here i go, idk if im right:
62 km/h = 223200 m/s
V2= 223200 m/s
a = 3.0 m/s^2
V1 = 0
t = ?
V2 = V1 + A*T
T = V2 - V1/A
T = 20.6 s
?

Answer 3

You got the right idea, but your conversion from km/h to m/s is incorrect.

Answer 4

sorry, its 17.2 m/s correct?

Answer 5

Yeah, that sounds right.

Answer 6

so the time is 5.7 seconds*, thanks! and there is part b and c...
b) calculate the speed of the police officer when he catches the speeder, is this reasonable?
c) now 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?

Answer 7

Hi, I just realized for the first part, you should find the distance at which the policeman catches the speeder. Use d = d_0 + v_0*t + 0/5*a*t^2 for both the police and speeder and set them equal. d_0 should be 0 for both. v_0 is 0 for the police. acceleration is 0 for the speeder because he's traveling at a constant speed. This should give you the correct time.

Answer 8

Police: d = d_0 + v_0*t + 0.5*a*t^2 (Set d_0 = 0 and v_0 = 0)
Speeder: d = d_0 + v_0*t + 0.5*a*t^2 (Set d_0 = 0 and a = 0)

This simplifies to
Police: d = 0.5*a*t^2
Speeder d = v_0*t

Since the speeder gets caught when he and the police have the same 'd', set them equal.

So 0.5*a*t^2 = v_0*t

0.5*3*t^2 = 17.2*t

t = 11.48 s

Answer 9

For part b), you know the police is accelerating at 3 m/s^2 for 11.48 seconds. His speed when he catches the speeder will be 3*11.48 = 34.44 m/s. This is about 77 mph, so this is reasonable.

Answer 10

thanks dude!!
i think i got part c from your response