# Kinematics Q

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#1

Question 14 here is proving extremely difficult and I'm becoming frustrated now as I just cant make out how to do it. Could anyone please give me some pointers?
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1 year ago
#2
(Original post by jamiet018)

Question 14 here is proving extremely difficult and I'm becoming frustrated now as I just cant make out how to do it. Could anyone please give me some pointers?
The bit you're missing is probably "just catches the bus". This implies that the bus and the man are going at the same speed when he catches up to it.

You should be able to form three equations:

1) Displacement of bus, in terms of the buses acceleration when he catches it.

2) Displacement of man in terms of his velocity.

3) Velocity of man = velocity of bus, when they meet.

That's enough to derive the desired quantities.
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1 year ago
#3
(Original post by jamiet018)
Question 14 here is proving extremely difficult and I'm becoming frustrated now as I just cant make out how to do it. Could anyone please give me some pointers?
If the bus is travelling from rest, starting 50m away from the man, with uniform acceleration , then its SUVAT equation for displacement is given by: .

Since the man is running at a constant speed, lets say it's , then his displacement is given by: .

These simplify to be:  .

30 seconds later, the man catches up to the bus. At this point, they both have the same displacement. Hence, must hold for . This yields you an equation in and .

You need another equation in and in order to solve these two simultaneously for acceleration of the bus and the speed of the man.

This other equation is not so obvious, but we know that the man starts running at some speed u and the bus has increasing speed starting at 0.
- While the speed of the bus is less than u, the man is catching up to the bus.
- When the speed of the bus is equal to u, the man is no longer catching up to the bus. They maintain a constant separation distance for a moment.
- Once the speed of the bus is greater than u, the distance between the man and the bus is increasing, as the bus is driving away.

We are told that the man *just* catches the bus. This means that the speed of the bus is < u when time is , and when they must obtain the same speed at the moment the man catches up to it.

This yields our second equation, speed of the bus at t=30 is equal to u. I.e. .

EDIT: Beaten to it 1
#4
(Original post by ghostwalker)
The bit you're missing is probably "just catches the bus". This implies that the bus and the man are going at the same speed when he catches up to it.

You should be able to form three equations:

1) Displacement of bus, in terms of the buses acceleration when he catches it.

2) Displacement of man in terms of his velocity.

3) Velocity of man = velocity of bus, when they meet.

That's enough to derive the desired quantities.
So I've done the first two (I think)
Bus. S = 30u + 450a
Man. S = 30v - 450a
Is this correct?
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1 year ago
#5
(Original post by jamiet018)
So I've done the first two (I think)
Bus. S = 30u + 450a
Man. S = 30v - 450a
Is this correct?
They don't look right, but you've not said what u and v are here, nor how you've derived them.

Have a look at RDKGames' post, as it's more detailed that mine.
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#6
(Original post by RDKGames)
If the bus is travelling from rest, starting 50m away from the man, with uniform acceleration , then its SUVAT equation for displacement is given by: .

Since the man is running at a constant speed, lets say it's , then his displacement is given by: .

These simplify to be:  .

30 seconds later, the man catches up to the bus. At this point, they both have the same displacement. Hence, must hold for . This yields you an equation in and .

You need another equation in and in order to solve these two simultaneously for acceleration of the bus and the speed of the man.

This other equation is not so obvious, but we know that the man starts running at some speed u and the bus has increasing speed starting at 0.
- While the speed of the bus is less than u, the man is catching up to the bus.
- When the speed of the bus is equal to u, the man is no longer catching up to the bus. They maintain a constant separation distance for a moment.
- Once the speed of the bus is greater than u, the distance between the man and the bus is increasing, as the bus is driving away.

We are told that the man *just* catches the bus. This means that the speed of the bus is < u when time is , and when they must obtain the same speed at the moment the man catches up to it.

This yields our second equation, speed of the bus at t=30 is equal to u. I.e. .

EDIT: Beaten to it Still really struggling to understand this I'm afraid. As a last resort is there any chance yourself (or any anybody else) could actually show me all the working out required. I know that isnt the way I should be going about it but been as I've probably spent about 3 hours now on this question it might be the only way I'm going to be able to understand
0
1 year ago
#7
(Original post by jamiet018)
Still really struggling to understand this I'm afraid. As a last resort is there any chance yourself (or any anybody else) could actually show me all the working out required. I know that isnt the way I should be going about it but been as I've probably spent about 3 hours now on this question it might be the only way I'm going to be able to understand
My post is like 75% of the working out right there. All you need to do is understand what I'm saying, then do the maths I point out.

Which bit are you struggling to grasp in my post?
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#8
(Original post by RDKGames)
My post is like 75% of the working out right there. All you need to do is understand what I'm saying, then do the maths I point out.

Which bit are you struggling to grasp in my post?
I will show you where I'm at:
50 1/2at^2 = Ut
Insert t=30
50 450a = 30u
5 45a = 3u
3u - 45a =5
Hopefully this is right so far. After this I'm stuck. I dont understand how to produce the next simultaneous equation.

Last edited by jamiet018; 1 year ago
0
1 year ago
#9
(Original post by jamiet018)
I will show you where I'm at:
50 1/2at^2 = Ut
Insert t=30
50 450a = 30u
5 45a = 3u
3u - 45a =5
Hopefully this is right so far. After this I'm stuck. I dont understand how to produce the next simultaneous equation.

That's right so far.

The other equation is given by , and I explain where this comes from in my first post. It's because at t=30 thats when the man *just* catches up to the bus, so the bus must obtain the same speed as the man at this time. The speed of the bus at is given by , so it's just at that moment. The speed of the man is still , hence the equality.
Last edited by RDKGames; 1 year ago
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#10
(Original post by RDKGames)
That's right so far.

The other equation is given by , and I explain where this comes from in my first post. It's because at t=30 thats when the man *just* catches up to the bus, so the bus must obtain the same speed as the man at this time. The speed of the bus at is given by , so it's just at that moment. The speed of the man is still , hence the equality.
Ah. Finally realised! 3u-1.5u=5 and so u=3.3 recurring. Thanks very much
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