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OCR G481 Physics A AS - Mechanics - 12-1-11

I looked in the Physics exam section and didn't see one for G481, so here it is :smile:

I'm a little bit stuck on an overtaking question from a past paper:

http://www.ocr.org.uk/download/pp_09_jan/ocr_34838_pp_09_jan_gce_g481_01.pdf

Question 1ciii, part 2
I can get the answer provided in the markscheme, 2.3 seconds, but what method do you use if the overtaking happens outside of the period that the car is accelerating? Surely s=ut+1/2at^2 wouldn't work, as the acceleration is zero?
Reply 1
Avatar for Jukeboxing
Jukeboxing
8
what you need to do is make the s=ut+1/2at^2 of A equal to the s=vt - 1/2at^2 of B.

A: s=10xt + 0.5x3.5xt^2 (ut + 0.5at^2)
B: s=14xt - 0.5x0xt^2 (vt - 0.5at^2)

A: s = 10t + 1.75t^2
B: s = 14t

So, make them equal to each other and you get this:

10t + 1.75t^2 = 14t
1.75t^2 = 14t - 10t
1.75t^2 = 4t (Now you can divide both sides by t or factorise, im going to divide by t)
1.75t = 4
t = 4/1.75
t = 2.2857....

I've got this exam aswell i'm doing all the past papers i can. If you want the June 2010 paper you get it here: http://kggsphysics.wikispaces.com/AS+full+past+papers+and+mks#old
(edited 13 years ago)
Reply 2
Avatar for wibletg
wibletg
OP
3
Original post by Jukeboxing
what you need to do is make the s=ut+1/2at^2 of A equal to the s=vt - 1/2at^2 of B.

A: s=10xt + 0.5x3.5xt^2 (ut + 0.5at^2)
B: s=14xt - 0.5x0xt^2 (vt - 0.5at^2)

A: s = 10t + 1.75t^2
B: s = 14t

So, make them equal to each other and you get this:

10t + 1.75t^2 = 14t
1.75t^2 = 14t - 10t
1.75t^2 = 4t (Now you can divide both sides by t or factorise, im going to divide by t)
1.75t = 4
t = 4/1.75
t = 2.2857....

I've got this exam aswell i'm doing all the past papers i can. If you want the June 2010 paper you get it here: http://kggsphysics.wikispaces.com/AS+full+past+papers+and+mks#old


Yeah mate, but my question is, how do you know that the overtaking takes place during the acceleration phase? What do you do if the overtaking takes place in the time when A has constant velocity?
Reply 3
Avatar for itzRussian
itzRussian
1
Original post by wibletg
Yeah mate, but my question is, how do you know that the overtaking takes place during the acceleration phase? What do you do if the overtaking takes place in the time when A has constant velocity?


Becauuuuuuse, assuming at t=0 the cars are side by side, you know that the displacement is given by the area under the graph. Car A overtakes car B just when its (under-graph) area becomes equal. The guy above derived equations for displacement, S for both cars. Since as we said before, areas will be equal when one overtakes the other, we put the equations side-by-side and solve for variable 't' which is what we want to find.
Reply 4
Avatar for wibletg
wibletg
OP
3
Original post by itzRussian
Becauuuuuuse, assuming at t=0 the cars are side by side, you know that the displacement is given by the area under the graph. Car A overtakes car B just when its (under-graph) area becomes equal. The guy above derived equations for displacement, S for both cars. Since as we said before, areas will be equal when one overtakes the other, we put the equations side-by-side and solve for variable 't' which is what we want to find.


Ah right, that makes sense, the question does say at t=0 the cars are side by side, cheers :smile:
Instead of making a thread on this, my question is why do we not use the whole equation of motion of object a?

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