Maths mechanics as question suvat

Original post by username79352

https://imgur.com/a/EYzmmFj
Can anyone help me with part b please? For part a I got t as 4 and 8 and for part b I think that the answer is 12 because from start back to a is 8 sec and then add four it takes for o to a. But I feel like this is wrong because it’s three marks. Could anyone explain whether my process is correct or whether I should be doing something else ?

Right answer, but for 3 marks theyd probably expect a suvat equation so one is
0 = 18t -3/2 t^2 = t/2 ( 36 - 3t) ....

As the displacement - time relationship is a parabola, its symmetric about the peak of 6s so will pass throgh 0m displacement at 0s and 12s.

(edited 1 month ago)

Reply 2

OP

Original post by mqb2766

Right answer, but for 3 marks theyd probably expect a suvat equation so one is
0 = 18t -3/2 t^2 = t/2 ( 36 - 3t) ....

For the first part of that equation, I did that in part but equated it to 48, to get 4 and 8. In the exam would they expect me to rewrite it all again with s as 0 or could I just use my part a answers to get the same answer? Would I lose marks for doing that if I have a brief explanation alongside?

Reply 3

Original post by username79352

For the first part of that equation, I did that in part but equated it to 48, to get 4 and 8. In the exam would they expect me to rewrite it all again with s as 0 or could I just use my part a answers to get the same answer? Would I lose marks for doing that if I have a brief explanation alongside?

Its hard to say, but its a fairly trivial factorisation for s=0 as t=0 is a trivial root and the other can be written down t=36/3. But if you properly justified it as a parabola with a peak at 6s (mean of 4s and 8s or complete the square) so by symmetry s=0 at 0 and 12, its gcse knowledge so should get the marks. If its an ad hoc justification, you may lose a mark or two.

(edited 1 month ago)

Reply 4

OP

Original post by mqb2766

Its hard to say, but its a fairly trivial factorisation for s=0 as t=0 is a trivial root and the other can be written down t=36/3. But if you properly justified it as a parabola with a peak at 6s (mean of 4s and 8s or complete the square) so by symmetry s=0 at 0 and 12, its gcse knowledge so should get the marks. If its an ad hoc justification, you may lose a mark or two.

My explanation was that it took four seconds to travel OA and the second time it was at A t was 8 so the total time would be time at A add time to travel distance OA or ao in this case which is 4 sec and added together they give 12

Reply 5

Original post by username79352

My explanation was that it took four seconds to travel OA and the second time it was at A t was 8 so the total time would be time at A add time to travel distance OA or ao in this case which is 4 sec and added together they give 12

You didnt mention its a parabola/quadratic relationship and the relevant/associated properties.

(edited 1 month ago)

Reply 6

OP

Original post by mqb2766

You didnt mention its a parabola/quadratic and the relevant/associated properties.

Ok so I’d lose marks then?

Reply 7

Original post by username79352

Ok so I’d lose marks then?

Possibly. Its obviously the right answer, but the explanation in the original post isnt that convincing. As in the first reply, its a trivial suvat factorisation to do it with an equation or note the properties of the parabolic displacement - time relationship to justify your explanation (or similar). Its probalby a similar time to do it either way.

Reply 8

OP