The Student Room Group

Objects in contact- what time do they meet

See attached:

6230213D-E7C6-4692-84A2-78B98F42BAD6.jpeg

Okay, so part a I found the force they move at is 20.8N hence acceleration of second skater is 0.306…

Part b.

They’re 1.2m apart. So, I need to calculate each skater separately to see which time they both meet the middle(0.6m)

Where am I going wrong?
First skater: u=0, s=0.6 , a=0.4 t=?

Using s=ut+½at²
t=√3

Second skater same above other than a=0.306
t=1.98 hence 3.71 seconds total for them to meet in the middle. Is my thinking correct?

I know one is accelerating faster than the other, so that would mean one travels more distance than the other would. Would this effect the outcome?
(edited 1 year ago)
Original post by KingRich
See attached:

Okay, so part a I found the force they move at is 20.8N hence acceleration of second skater is 0.306…

Part b.

They’re 1.2m apart. So, I need to calculate each skater separately to see which time they both meet the middle(0.6m)

Where am I going wrong?
First skater: u=0, s=0.6 , a=0.4 t=?

Using s=ut+½at²
t=√3

Second skater same above other than a=0.306
t=1.98 hence 3.71 seconds total for them to meet in the middle. Is my thinking correct?

I know one is accelerating faster than the other, so that would mean one travels more distance than the other would. Would this effect the outcome?


Attached? However (guessing the question), if they have different accelerations, they wont meet in the middle. Let one travel a distance s, then the other travels 1.2-s and you want to find the time when this occurs. But make sure you get signs etc right.
Reply 2
Original post by mqb2766
Attached? However (guessing the question), if they have different accelerations, they wont meet in the middle. Let one travel a distance s, then the other travels 1.2-s and you want to find the time when this occurs. But make sure you get signs etc right.


Apologies. Mmm, If it’s as I was thinking, it’s something along the lines of if the fastest accelerating one travels 1.2m in so so time.

then the one that accelerates slower would travel less distance..

I just wasn’t sure how to word it. I’ll have a go
Original post by KingRich
Apologies. Mmm, If it’s as I was thinking, it’s something along the lines of if the fastest accelerating one travels 1.2m in so so time.

then the one that accelerates slower would travel less distance..

I just wasn’t sure how to word it. I’ll have a go

Agree with the acceleration of the second skater. There is a quicker way to do it using relative motion, but have a go as per above first.

Edit - note in the OP you add the times. This isn't correct as you want them to come together at the same time.
(edited 1 year ago)
Reply 4
Original post by mqb2766
Agree with the acceleration of the second skater. There is a quicker way to do it using relative motion, but have a go as per above first.


I assume that’s something that I do with projected acceleration of both? I’m not sure, I’m just guessing.

so, for the fastest accelerating skater they travel 1.2m in 2.45 seconds.

If in that time the second skater at acceleration 0.306 it covers s=0.88128metres

hence, 1.2-0.88128=0.31872

If this is the moment in meters that they meet. Do I calculate the faster accelerating skater to find t?

Edit: I should calculate for both skaters and then add that time I believe
(edited 1 year ago)
Original post by KingRich
I assume that’s something that I do with projected acceleration of both? I’m not sure, I’m just guessing.

so, for the fastest accelerating skater they travel 1.2m in 2.45 seconds.

If in that time the second skater at acceleration 0.306 it covers s=0.88128metres

hence, 1.2-0.88128=0.31872

If this is the moment in meters that they meet. Do I calculate the faster accelerating skater to find t?

Not sure what youre doing. You need to set up
first skater: s = ut + 1/2 at^2
second skater 1.2-s = ut + 1/2 at^2
where obviously the accelerations are different. Then solve simultaneously for s and t so that at that time, they have moved a distance of 1.2m in total. s is the distance moved by skater 1, and 1.2-s by skater 2, so 1.2 in total.
Reply 6
Original post by mqb2766
Not sure what youre doing. You need to set up
first skater: s = ut + 1/2 at^2
second skater 1.2-s = ut + 1/2 at^2
where obviously the accelerations are different. Then solve simultaneously for s and t so that at that time, they have moved a distance of 1.2m in total. s is the distance moved by skater 1, and 1.2-s by skater 2, so 1.2 in total.


Erm, if I set up the first skater at 1.2m then it would have already travelled 1.2 by itself. Do I set this up as 1.2-s too?

6168EEA8-40E7-4DDA-972C-77280CFE1541.jpeg

This is what I’ve done

Edit: so, skater two travels 0.282 metres and skater one travels 0.918m. It’s at this point they both meet. As that’s a total of 1.2m.
That’s my understanding so far based on the calculations

Okay. Got there in the end!!!

553B7C85-5637-4ECD-8CD9-7B93655252FD.jpeg
(edited 1 year ago)
Edit - solve the simultaneous equations
first skater: s = ut + 1/2 0.4 t^2
second skater 1.2-s = ut + 1/2 0.308 t^2
The question only asks for the time, so sub the first equation into the second and "solve the quadratic" for t.

Didnt see your edit, so yes for the final attempt. Note that it should be a couple of lines if you think of the relative motion. Assuming skater 1 is fixed (motion is relative to them), then the relative acceleration of skater 2 is 0.4+0.308=0.708 (you can do this using exact arithmetic) and skater 2 will travel a distance of 1.2 relative to skater 1 so
1.2 = 1/2 0.708 t^2
and solving gives t = sqrt(17/5) ~ 1.84. When you sub the first equation into the second, this is what you get. So setting it up as a relative motion problem saves setting up the two equations and reducing them to a single equation which represents the relative motion.
(edited 1 year ago)
Reply 8
Original post by mqb2766
Edit - solve the simultaneous equations
first skater: s = ut + 1/2 0.4 t^2
second skater 1.2-s = ut + 1/2 0.308 t^2
The question only asks for the time, so sub the first equation into the second and "solve the quadratic" for t.

Didnt see your edit, so yes for the final attempt. Note that it should be a couple of lines if you think of the relative motion. Assuming skater 1 is fixed (motion is relative to them), then the relative acceleration of skater 2 is 0.4+0.308=0.708 (you can do this using exact arithmetic) and skater 2 will travel a distance of 1.2 relative to skater 1 so
1.2 = 1/2 0.708 t^2
and solving gives t = sqrt(17/5) ~ 1.84. When you sub the first equation into the second, this is what you get. So setting it up as a relative motion problem saves setting up the two equations and reducing them to a single equation which represents the relative motion.


Mmm, so I was correct in the assumption of doing something with both their accelerations. So, adding them both and equating to the distance required.

applying s=ut+½at²

based on the sum of both their accelerations. That’s extremely useful and far more direct. Mmm, I don’t believe I’ve touched on that concept yet. Unless that’s year two where they delve deeper into the section.

what is this concept called? I’ll write it down
Original post by KingRich
Mmm, so I was correct in the assumption of doing something with both their accelerations. So, adding them both and equating to the distance required.

applying s=ut+½at²

based on the sum of both their accelerations. That’s extremely useful and far more direct. Mmm, I don’t believe I’ve touched on that concept yet. Unless that’s year two where they delve deeper into the section.

what is this concept called? I’ll write it down

Its just relative motion, so the motion of the second body relative to the first. So think about cars travelling in opposite directions and youre in one car as opposed to watching both cars from side of the road. As far as I know, its not really on the a level syllabus, but questions of this type are often simplified if you approach them like this.

Note that it falls out of the setting up two simultaneous equations then subbing one into the other. You either do relative motion by setting up the problem like that or it arises as part of the fiddling around with the algebra, even if you don't recognise that its happening.

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