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#1
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

0
2 years ago
#2
(Original post by 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

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.
0
2 years ago
#3
(Original post by 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

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!
0
2 years ago
#4
(Original post by 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!
For the second part, you could find 2 expressions in t for the displacement of the man and the bus.
Let D= difference between them, then solve dD/dt = 0 to determine max possible gap.
0
2 years ago
#5
(Original post by 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!
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".
1
2 years ago
#6
(Original post by 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 velocity of man was greater than velocity of bus then it wouldn't make sense to say he "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.
0
2 years ago
#7
S = ut + 1/2at^2
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#8
(Original post by 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!
Even if that's the case, I still can't get an answer.
0
#9
(Original post by 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.
Ok, thanks.
0
2 years ago
#10
(Original post by Al4stair)
Even if that's the case, I still can't get an answer.
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
1
#11
(Original post by 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
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 (:
0
#12
(Original post by 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".
Yes, totally agree. The question is worded badly in my opinion and I would have never have tried it the way vc94 has. 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.
0
2 years ago
#13
(Original post by 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 (:
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
0
#14
(Original post by 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
The question only says the man's speed has changed, not the speed of the bus. Ignoring that, I understand what you have done up until 'max possible gap we must have dD/dt = 0', I don't what you're doing there and what that means. And surely we want the smallest gap, since the question asks 'how close to the bus can he get'?
0
#15
(Original post by 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
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?
0
2 years ago
#16
(Original post by 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?
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
0
#17
(Original post by 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
I have spent ages to try and get to Vb=1/9 but can't get it. How did you go from Sb= 50 + (1/18)t^2 to Vb=1/9??
0
#18
(Original post by 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
I get this though. Here's my working: https://scontent.fbhx2-1.fna.fbcdn.n...7c&oe=5BC10FF6
0
#19
(Original post by 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
I am still unsure on why you are referring to the gap as the 'max possible gap' since we want to find how close the man gets to the bus, not how far away he gets. I am not sure if you are describing something different to what I'm thinking since you're talking about the speed of man being different to that of the final speed of the bus.
"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.
0
#20
(Original post by 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
In case you didn't see the edits, I completely understand what you mean. Thanks 0
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