# Kinematics Q

Watch
Announcements

Page 1 of 1

Go to first unread

Skip to page:

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?

0

reply

Report

#2

(Original post by

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?

**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?

*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.

0

reply

Report

#3

**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?

.

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

reply

(Original post by

The bit you're missing is probably "

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.

**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.

Bus. S = 30u + 450a

Man. S = 30v - 450a

Is this correct?

0

reply

Report

#5

(Original post by

So I've done the first two (I think)

Bus. S = 30u + 450a

Man. S = 30v - 450a

Is this correct?

**jamiet018**)So I've done the first two (I think)

Bus. S = 30u + 450a

Man. S = 30v - 450a

Is this correct?

Have a look at RDKGames' post, as it's more detailed that mine.

0

reply

(Original post by

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

**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

0

reply

Report

#7

(Original post by

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

**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

Which bit are you struggling to grasp in my post?

0

reply

(Original post by

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?

**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?

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.

Thanks for your patience btw

Last edited by jamiet018; 1 year ago

0

reply

Report

#9

(Original post by

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.

Thanks for your patience btw

**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.

Thanks for your patience btw

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

0

reply

(Original post by

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.

**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.

0

reply

X

Page 1 of 1

Go to first unread

Skip to page:

### Quick Reply

Back

to top

to top