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Please can someone show me how to solve this mechanics question (question 16)?:

https://scontent.fbhx2-1.fna.fbcdn.n...d5&oe=5BC052A6

I don't understand the wording of the question and there doesn't seem to be enough info to answer it.

Attempt 1:https://scontent.fbhx2-1.fna.fbcdn.n...23&oe=5B7A01C0

Attempt 2:https://scontent.fbhx2-1.fna.fbcdn.n...c5&oe=5B826DD1

Textbook answer:https://scontent.xx.fbcdn.net/v/t1.1...52&oe=5B813517

https://scontent.fbhx2-1.fna.fbcdn.n...d5&oe=5BC052A6

I don't understand the wording of the question and there doesn't seem to be enough info to answer it.

Attempt 1:https://scontent.fbhx2-1.fna.fbcdn.n...23&oe=5B7A01C0

Attempt 2:https://scontent.fbhx2-1.fna.fbcdn.n...c5&oe=5B826DD1

Textbook answer:https://scontent.xx.fbcdn.net/v/t1.1...52&oe=5B813517

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#2

(Original post by

Please can someone show me how to solve this mechanics question (question 16)?:

https://scontent.fbhx2-1.fna.fbcdn.n...d5&oe=5BC052A6

I don't understand the wording of the question and there doesn't seem to be enough info to answer it.

Attempt 1:https://scontent.fbhx2-1.fna.fbcdn.n...23&oe=5B7A01C0

Attempt 2:https://scontent.fbhx2-1.fna.fbcdn.n...c5&oe=5B826DD1

Textbook answer:https://scontent.xx.fbcdn.net/v/t1.1...52&oe=5B813517

**Al4stair**)Please can someone show me how to solve this mechanics question (question 16)?:

https://scontent.fbhx2-1.fna.fbcdn.n...d5&oe=5BC052A6

I don't understand the wording of the question and there doesn't seem to be enough info to answer it.

Attempt 1:https://scontent.fbhx2-1.fna.fbcdn.n...23&oe=5B7A01C0

Attempt 2:https://scontent.fbhx2-1.fna.fbcdn.n...c5&oe=5B826DD1

Textbook answer:https://scontent.xx.fbcdn.net/v/t1.1...52&oe=5B813517

**from rest**), so if the bus's acceleration is a, it travels 1/2at^2 metres in t seconds, and thus as it started at x=50, its position at time t is (50 + 1/2at^2). Now for the man, he runs at a constant speed - call this speed v, so that his position at time t is vt. Now when do the man and the bus meet? The answer is when 50 + 1/2at^2 = vt, and we need this to be satisfied when t = 30, so 50 + 450a = 30t, so 5 + 45a = 3t. We have now used all the information and have only one equation in two variables, so the conclusion is that

**the question is unsolvable**.

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#3

**Al4stair**)

Please can someone show me how to solve this mechanics question (question 16)?:

https://scontent.fbhx2-1.fna.fbcdn.n...d5&oe=5BC052A6

I don't understand the wording of the question and there doesn't seem to be enough info to answer it.

Attempt 1:https://scontent.fbhx2-1.fna.fbcdn.n...23&oe=5B7A01C0

Attempt 2:https://scontent.fbhx2-1.fna.fbcdn.n...c5&oe=5B826DD1

Textbook answer:https://scontent.xx.fbcdn.net/v/t1.1...52&oe=5B813517

The correct values come out!

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#4

(Original post by

When the question says "just catches the bus in 30s" then you can assume that the velocity of man = velocity of bus when t=30.

The correct values come out!

**vc94**)When the question says "just catches the bus in 30s" then you can assume that the velocity of man = velocity of bus when t=30.

The correct values come out!

Let D= difference between them, then solve dD/dt = 0 to determine max possible gap.

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#5

**vc94**)

When the question says "just catches the bus in 30s" then you can assume that the velocity of man = velocity of bus when t=30.

The correct values come out!

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#6

(Original post by

Hmm OK. Then this seems more like an English problem than a maths problem, as neither I nor (presumably) the OP inferred this from the phrase "just catches the bus".

**Prasiortle**)Hmm OK. Then this seems more like an English problem than a maths problem, as neither I nor (presumably) the OP inferred this from the phrase "just catches the bus".

If the velocity of the man is less than the velocity of bus then he wouldn't catch up with it at all.

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**vc94**)

When the question says "just catches the bus in 30s" then you can assume that the velocity of man = velocity of bus when t=30.

The correct values come out!

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(Original post by

Let x represent position as a function of time, with x=0 representing the man's starting point, and thus x=50 representing the bus stop. Using s=ut+1/2at^2, we know u = 0 (accelerating

**Prasiortle**)Let x represent position as a function of time, with x=0 representing the man's starting point, and thus x=50 representing the bus stop. Using s=ut+1/2at^2, we know u = 0 (accelerating

**from rest**), so if the bus's acceleration is a, it travels 1/2at^2 metres in t seconds, and thus as it started at x=50, its position at time t is (50 + 1/2at^2). Now for the man, he runs at a constant speed - call this speed v, so that his position at time t is vt. Now when do the man and the bus meet? The answer is when 50 + 1/2at^2 = vt, and we need this to be satisfied when t = 30, so 50 + 450a = 30t, so 5 + 45a = 3t. We have now used all the information and have only one equation in two variables, so the conclusion is that**the question is unsolvable**.
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#10

(Original post by

Even if that's the case, I still can't get an answer.

**Al4stair**)Even if that's the case, I still can't get an answer.

Displacement of man: Sm= 30x

Displacement of bus: Sb= 50+ vt -0.5at^2 = 50 + 30x -0.5a(30^2)

When the man catches the bus, Sm=Sb, solving will give

**a=1/9**

Also, Displacement of bus: Sb= 50 + ut +0.5at^2 = 50 + 0.5(1/9)*30^2 = 100

So Sm =30x=100 gives

**x=10/3 m/s**

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(Original post by

Let Vm=velocity of man and Vb=velocity of bus. When the man "just catches" the bus, Vm=Vb = x and t=30.

Displacement of man: Sm= 30x

Displacement of bus: Sb= 50+ vt -0.5at^2 = 50 + 30x -0.5a(30^2)

When the man catches the bus, Sm=Sb, solving will give

Also, Displacement of bus: Sb= 50 + ut +0.5at^2 = 50 + 0.5(1/9)*30^2 = 100

So Sm =30x=100 gives

**vc94**)Let Vm=velocity of man and Vb=velocity of bus. When the man "just catches" the bus, Vm=Vb = x and t=30.

Displacement of man: Sm= 30x

Displacement of bus: Sb= 50+ vt -0.5at^2 = 50 + 30x -0.5a(30^2)

When the man catches the bus, Sm=Sb, solving will give

**a=1/9**Also, Displacement of bus: Sb= 50 + ut +0.5at^2 = 50 + 0.5(1/9)*30^2 = 100

So Sm =30x=100 gives

**x=10/3 m/s**The question is worded badly in my opinion and I would have never have tried it that way. It poorly tries to imply the end velocity of the man and the bus is the same. You probably wouldn't jump on a moving bus because the doors would be closed and the in this case the bus is travelling at 7 mph, pretty fast for someone to jump on. Therefore the question could be interpreted in many different ways. It says he 'catches' the bus, this could be interpreted as he catches the bus in speed, for example his final speed might by 8 mph, and then he caught up the bus (and overtook it) and was then ran over.

But I am having issues getting the same answer as the book for 'if the man's speed was 3ms^-1'. I get 10m, but the book get 9.5m (I didn't round). Do you know why this is? My workings for that part: https://scontent.fbhx2-1.fna.fbcdn.n...0d&oe=5BBF4B22

Working for the acceleration of the bus: https://scontent.fbhx2-1.fna.fbcdn.n...91&oe=5B7FF660

Anyway, thanks again (:

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**Prasiortle**)

Hmm OK. Then this seems more like an English problem than a maths problem, as neither I nor (presumably) the OP inferred this from the phrase "just catches the bus".

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#13

(Original post by

Wow. Thanks a lot! I checked through all that and it yeah I get the same.

The question is worded badly in my opinion and I would have never have tried it that way. It poorly tries to imply the end velocity of the man and the bus is the same. You probably wouldn't jump on a moving bus because the doors would be closed and the in this case the bus is travelling at 7 mph, pretty fast for someone to jump on. Therefore the question could be interpreted in many different ways. It says he 'catches' the bus, this could be interpreted as he catches the bus in speed, for example his final speed might by 8 mph, and then he caught up the bus (and overtook it) and was then ran over.

But I am having issues getting the same answer as the book for 'if the man's speed was 3ms^-1'. I get 10m, but the book get 9.5m (I didn't round). Do you know why this is? My workings for that part: https://scontent.fbhx2-1.fna.fbcdn.n...0d&oe=5BBF4B22

Working for the acceleration of the bus: https://scontent.fbhx2-1.fna.fbcdn.n...91&oe=5B7FF660

Anyway, thanks again (:

**Al4stair**)Wow. Thanks a lot! I checked through all that and it yeah I get the same.

The question is worded badly in my opinion and I would have never have tried it that way. It poorly tries to imply the end velocity of the man and the bus is the same. You probably wouldn't jump on a moving bus because the doors would be closed and the in this case the bus is travelling at 7 mph, pretty fast for someone to jump on. Therefore the question could be interpreted in many different ways. It says he 'catches' the bus, this could be interpreted as he catches the bus in speed, for example his final speed might by 8 mph, and then he caught up the bus (and overtook it) and was then ran over.

But I am having issues getting the same answer as the book for 'if the man's speed was 3ms^-1'. I get 10m, but the book get 9.5m (I didn't round). Do you know why this is? My workings for that part: https://scontent.fbhx2-1.fna.fbcdn.n...0d&oe=5BBF4B22

Working for the acceleration of the bus: https://scontent.fbhx2-1.fna.fbcdn.n...91&oe=5B7FF660

Anyway, thanks again (:

For the greatest gap: Vm= 3m/s =Vb

So

**Sm=3t**and Sb=50 +ut + 0.5a(t^2) = 50 + 0.5*(1/9)t^2

So

**Sb= 50 + (1/18)t^2**

Let

**D= Sb - Sm = 50 +(1/18)t^2 -3t**then for the max possible gap we must have dD/dt = 0 ...which gives t/9 - 3 =0, so t=27 sec

Hence D= 50 + (27^2)/18 - 3*27 =

**9.5 m**

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(Original post by

Here's what I did for the second part. The wording in this question is a trainwreck!

For the greatest gap: Vm= 3m/s =Vb

So

So

Let

Hence D= 50 + (27^2)/18 - 3*27 =

**vc94**)Here's what I did for the second part. The wording in this question is a trainwreck!

For the greatest gap: Vm= 3m/s =Vb

So

**Sm=3t**and Sb=50 +ut + 0.5a(t^2) = 50 + 0.5*(1/9)t^2So

**Sb= 50 + (1/18)t^2**Let

**D= Sb - Sm = 50 +(1/18)t^2 -3t**then for the max possible gap we must have dD/dt = 0 ...which gives t/9 - 3 =0, so t=27 secHence D= 50 + (27^2)/18 - 3*27 =

**9.5 m**
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**vc94**)

Here's what I did for the second part. The wording in this question is a trainwreck!

For the greatest gap: Vm= 3m/s =Vb

So

**Sm=3t**and Sb=50 +ut + 0.5a(t^2) = 50 + 0.5*(1/9)t^2

So

**Sb= 50 + (1/18)t^2**

Let

**D= Sb - Sm = 50 +(1/18)t^2 -3t**then for the max possible gap we must have dD/dt = 0 ...which gives t/9 - 3 =0, so t=27 sec

Hence D= 50 + (27^2)/18 - 3*27 =

**9.5 m**

t=27 at the minimum point of the graph, so surely that's the minimum possible gap?

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#16

(Original post by

Here's the graph of what you did: https://www.desmos.com/calculator/kft0zjla0z

t=27 at the minimum point of the graph, so surely that's the minimum possible gap?

**Al4stair**)Here's the graph of what you did: https://www.desmos.com/calculator/kft0zjla0z

t=27 at the minimum point of the graph, so surely that's the minimum possible gap?

**Vm = Vb**.

You can't have Vm > Vb since this would contradict the man "just" catching the bus.

You can't have Vm < Vb since the man would never catch up with the bus!

Sm=3t means that Vm= 3

and Sb= 50 + (1/18)t^2 means that Vb= t/9.

So Vm=Vb gives t=27

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(Original post by

Another way to think about it is that the max possible gap is when

You can't have Vm > Vb since this would contradict the man "just" catching the bus.

You can't have Vm < Vb since the man would never catch up with the bus!

Sm=3t means that Vm= 3

and Sb= 50 + (1/18)t^2 means that Vb= t/9.

So Vm=Vb gives t=27

**vc94**)Another way to think about it is that the max possible gap is when

**Vm = Vb**.You can't have Vm > Vb since this would contradict the man "just" catching the bus.

You can't have Vm < Vb since the man would never catch up with the bus!

Sm=3t means that Vm= 3

and Sb= 50 + (1/18)t^2 means that Vb= t/9.

So Vm=Vb gives t=27

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**vc94**)

Here's what I did for the second part. The wording in this question is a trainwreck!

For the greatest gap: Vm= 3m/s =Vb

So

**Sm=3t**and Sb=50 +ut + 0.5a(t^2) = 50 + 0.5*(1/9)t^2

So

**Sb= 50 + (1/18)t^2**

Let

**D= Sb - Sm = 50 +(1/18)t^2 -3t**then for the max possible gap we must have dD/dt = 0 ...which gives t/9 - 3 =0, so t=27 sec

Hence D= 50 + (27^2)/18 - 3*27 =

**9.5 m**

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**vc94**)

Another way to think about it is that the max possible gap is when

**Vm = Vb**.

You can't have Vm > Vb since this would contradict the man "just" catching the bus.

You can't have Vm < Vb since the man would never catch up with the bus!

Sm=3t means that Vm= 3

and Sb= 50 + (1/18)t^2 means that Vb= t/9.

So Vm=Vb gives t=27

"You can't have Vm > Vb since this would contradict the man "just" catching the bus."- The man doesn't catch up to the bus when his speed his 3ms^-1 though?

"You can't have Vm < Vb since the man would never catch up with the bus!"- I thought he doesn't catch the bus in the case and the closest he gets to it is 9.5m?

Sorry for all these questions. I do really do appreciate your help.

EDIT: OMG. Oops! I completely misunderstood what you wrote. I feel like an idiot. Yes I completely understand what you mean about how the different possible final speeds and how they must equal each other. Yeah. v=u+at, 3=1/9 * t. t=27. And then using Sb= 50 + (1/18)t^2 you get 90.5m for the bus and sm=3t which gives you 81 for the man. 90.5-81 gives you 9.5m!!! Sorry for asking such dumb questions. Thanks again!!!!!!!!!!!!!! It makes perfect sense.

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**vc94**)

Another way to think about it is that the max possible gap is when

**Vm = Vb**.

You can't have Vm > Vb since this would contradict the man "just" catching the bus.

You can't have Vm < Vb since the man would never catch up with the bus!

Sm=3t means that Vm= 3

and Sb= 50 + (1/18)t^2 means that Vb= t/9.

So Vm=Vb gives t=27

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