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Urgent mechanics question!!

Hi, please could i have help on this question? I’m really confused why for part b, once I’ve integrated, to find c for the second part the markscheme says that at t=5s, the distance traveled is 0m?
Here is the question https://app.gemoo.com/share/image-annotation/639535308074926080?codeId=P570gKp5zmNQa&origin=imageurlgenerator

Thank you so much!!
Reply 1
Original post by anonymous294
Hi, please could i have help on this question? I’m really confused why for part b, once I’ve integrated, to find c for the second part the markscheme says that at t=5s, the distance traveled is 0m?
Here is the question https://app.gemoo.com/share/image-annotation/639535308074926080?codeId=P570gKp5zmNQa&origin=imageurlgenerator
Thank you so much!!

Presume its an initial condition (s(5)=0) for motion for the second part, but it would help to see the working youre referring to. So its probably simply saying thats the origin for the t>5 phase.

Its worth noting that the second phase is constant acceleration/suvat motion so there is no need to integrate.
(edited 1 month ago)
Original post by mqb2766
Presume its an initial condition (s(5)=0) for motion for the second part, but it would help to see the working youre referring to. So its probably simply saying thats the origin for the t>5 phase.
Its worth noting that the second phase is constant acceleration/suvat motion so there is no need to integrate.

This is what the markscheme says:
https://app.gemoo.com/share/image-annotation/639596129824014336?codeId=MpOKYRQ7ZXArV&origin=imageurlgenerator

But i am confused why they are substituting 10 seconds in the second part, i thought it travels 5 seconds in the first part and 5 in the next so we need to substitute 5 instead?
Thank you!
(edited 1 month ago)
t is not the time duration, it is the time passed. You can solve the question by introducing limits to the integral for the second interval. these limits would be 5 and 10
Original post by WordsFiddle
t is not the time duration, it is the time passed. You can solve the question by introducing limits to the integral for the second interval. these limits would be 5 and 10

similarly for the first interval the limits are 0 and 5.
Reply 5
Original post by anonymous294
This is what the markscheme says:
https://app.gemoo.com/share/image-annotation/639596129824014336?codeId=MpOKYRQ7ZXArV&origin=imageurlgenerator
But i am confused why they are substituting 10 seconds in the second part, i thought it travels 5 seconds in the first part and 5 in the next so we need to substitute 5 instead?
Thank you!

Youve really just got to work through the maths/sketch stuff. They break it at 5s as for t<5 its a positive displacement (relative to an origin when t=0) and for t>5 its a negative dispacement (relative to an origin when t=5) as the velocity is positive and then negative. For the second phase, you start at t=5 with u=0 and finish at t=10 with v=-10. a=-2. You could use that info with a suvat for S (negative displacement for the second phase) with T=5 (using uppercase just to try and keep things clear), or integrate velocity from t=5 to 10 with s=0 to S. They do the latter.
(edited 1 month ago)
Reply 6
Original post by anonymous294
This is what the markscheme says:
https://app.gemoo.com/share/image-annotation/639596129824014336?codeId=MpOKYRQ7ZXArV&origin=imageurlgenerator
But i am confused why they are substituting 10 seconds in the second part, i thought it travels 5 seconds in the first part and 5 in the next so we need to substitute 5 instead?
Thank you!

Note the "mark scheme" is overly complex for the second phase. It lasts T=5s and the average velocity is -5m/s so S=-25m.

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