Physics questions - kinematics

bonkysleuth SG

I've got 2 questions which need to be addressed urgently, so please help me out if possible.

1. A police car is at rest alongside a road monitoring passing cars when one passes at a constant speed of 32 m s-1. 5 seconds later, the police car accelerates from rest with an acceleration of 0.80 m s-2.

A)If the police car can maintain this acceleration, how far will it move before it draws level with the car?

B)What is the speed of the police car then?

2 A wooden crate slides from rest down a smooth inclined plane which is 5.0 m long and inclined at 30deg to the horizontal.

(a) find its acceleration down the plane

(b) calculate time taken for it to reach the bottom of the plane

the crate is now projected up the plane with an inital velocity of 6.0ms-1

(d) how long does the crate take to come to rest

lastly, thanks.

eagle SG

Is this A levels or O levels? The values to use and methods are very different for both.

1a) Let time = t

Draw your speed time graph to see what I mean when calculating distance

Distance speeding car travelled = 32t
Distance police car travelled = 0.5 * 0.80 * t^2

Equate them
0.4t^2 = 32t
Since t is not zero, 0.4t = 32
t = 80s

b) v = u + at
v = 0 + 0.8 * 80
v = 64 ms-1

Q2 I shall use the A levels method

2a) 9.81 sin 30 degrees= 4.9 ms-2

b) s = ut + 0.5 at^2,
5 = 0 + 0.5 * 4.9 * t^2
t^2 = 2.55
t = 1.6s

c) v^2 = u^2 + 2as
0 = 36 - 2 * 4.9 * s
9.8 s = 36
s = 3.7 m

d) v = u + at
0 = 6 - 4.9 t
4.9 t = 6
t = 1.22s

bonkysleuth SG

Is this A levels or O levels? The values to use and methods are very different for both.

1a) Let time = t

Draw your speed time graph to see what I mean when calculating distance

Distance speeding car travelled = 32t
Distance police car travelled = 0.5 * 0.80 * t^2

Equate them
0.4t^2 = 32t
Since t is not zero, 0.4t = 32
t = 80s

b) v = u + at
v = 0 + 0.8 * 80
v = 64 ms-1

Q2 I shall use the A levels method

2a) 5 sin 30 degrees= 2.5ms-2

b) s = ut + 0.5 at^2,
5 = 0 + 0.5 * 2.5 * t^2
t^2 = 4
t = 2s

c) v^2 = u^2 + 2as
0 = 36 - 2 * 2.5 * s
5s = 36
s = 7.2 m

d) v = u + at
0 = 6 - 2.5 t
2.5 t = 6
t = 2.4s

Hello, is there a prob with question 2? My teacher gave us the answers as 4.9ms-2, 1.4s, 3.7m and 1.2s. I do have a huge difficulty in understanding how to find the acceleration down the plane. Thanks for helpin anyways

eagle SG

oh ya, I used the wrong constants... Will edit now...

If u noticed, I accidentally used 5 instead of 9.81 for gravity...

bonkysleuth SG

Hi again... regarding the wooden crate on the plane question, for part (c) where they say the crate is projected up the plane, is it a projectile motion question? I don't think so because I do have friends who say it is. What does it actually mean by "projected up"? You mean throwing the box upwards after it has reached the bottom of the inclined plane? And why do we use a = 4.905ms-2 for calculations in (c) and (d) instead of 9.81ms-2? (for me, i'd have thought of using 9.81ms-2 since it is projected upwards and not along the slope) But I think my concept is wrong and needs clarification. thanks

eagle SG

Hmmm.... You use 4.9 in (c) and (d) because that's the acceleration of the crate down the plane as calculated in (a).

Remember that we are considering the kinematics along the line of the plane. The crane is not projected upwards; your question states that it is projected up the plane. It is along the slope.