Question on Equation Of Motion

Hello,

I have just started my as level and now ineed of help with this question. Please help me if you can.

A police car stationary on the side of the road sees a car passing at a speed of 40m/s. The police car immediately gives chase and accelerates at 3m/s^2 for 16s, followed by a contant speed.
(a) How long does it take for the police car to catch up the speeding motorist?
(b) What distance will the police car have travelled?

My attemp:

Police car:
initial velocity=0m/s
final velocity=?
acceleration=3m/s^2
time=16s

using V=u+at

V=48m/s

So, police car going at 48m/s and the other at 40m/s.

48t=40t
t=8s

They both trvelled for 8 seconds while travelling at 48 and 40.

Distace covered in 8 seconds is 384m.

So it takes 16s+8s=24s for the police car to catch the speeding motorist.

(b) Police car travelled 384m in 8 seconds and 384 while accelerating, so distacne travlled is 384+384=768m.

Is this correct, if not can you please tell me where i have gone wrong?

Thanks

Charikaar

Reply 1

sig.fig.

2

Umm, my M1 is a bit rusty but I got:

(a)

POLICE:

u = 0ms^-1 a = 3ms^-2 t = 16s

v = u + at = 0 + (3x16) = 48ms^-1

s = ut + 1/2at²
= 0 + (1/2 x 3 x 16² )
= 384M

CAR:

[for uniform velocity] s = vt = 40 x 16 = 640M

therefore, after 16s:

POLICE: s = 384M v = 48ms^-1
CAR: s = 640M v = 40ms^-1

When the Police catch up with the Car, s_police = s_car

Hence; v_pt + 384 = v_ct + 640
48t + 384 = 40t + 640
8t = 256
t = 32s
ergo, t_total = 32 + 16 = 48s

(b) 's' whilst accelerating = 384M
's' at constant velocity = vt = (48 x 32) = 1536
therefore, s_total = 1536 + 384 = 1920M

It's a hell of a lot easier though to draw a velocity-time graph and use the fact that the area under the graph = s. :biggrin:

Reply 2

charikaar

OP

trunksss6

Umm, my M1 is a bit rusty but I got:

(a)

POLICE:

u = 0ms^-1 a = 3ms^-2 t = 16s

v = u + at = 0 + (3x16) = 48ms^-1

s = ut + 1/2at²
= 0 + (1/2 x 3 x 16² )
= 384M

CAR:

[for uniform velocity] s = vt = 40 x 16 = 640M

therefore, after 16s:

POLICE: s = 384M v = 48ms^-1
CAR: s = 640M v = 40ms^-1

When the Police catch up with the Car, s_police = s_car

Hence; v_pt + 384 = v_ct + 640
48t + 384 = 40t + 640
8t = 256
t = 32s
ergo, t_total = 32 + 16 = 48s

(b) 's' whilst accelerating = 384M
's' at constant velocity = vt = (48 x 32) = 1536
therefore, s_total = 1536 + 384 = 1920M

It's a hell of a lot easier though to draw a velocity-time graph and use the fact that the area under the graph = s. :biggrin:

NICELY DONE!! THANK YOU VERY VERY MUCH.

Reply 3

sig.fig.

2

No probs.... (hope it's right) :p:

Question on Equation Of Motion

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