Can anyone help with this SUVAT A-level question?

amaniix

Which part / what have you tried?

(edited 4 months ago)

Reply 2

amaniix

OP

3

Original post by mqb2766

Which part / what have you tried?

I tried A by substituting t = 0 into t^2-0.4t-4.4 which gave me v = -4.4. And then I thought I can use v=u+at to find the initial acceleration. But do I sub u = 0 in for that? I'm not sure since then I'd be assuming it is at rest but the question doesn't state that. And if I did sub in u=0 then I'd get -4.4 = 0 +0a which obviously i can't do because you can't divide by 0

Reply 3

mqb2766

21

Original post by amaniix

I tried A by substituting t = 0 into t^2-0.4t-4.4 which gave me v = -4.4. And then I thought I can use v=u+at to find the initial acceleration. But do I sub u = 0 in for that? I'm not sure since then I'd be assuming it is at rest but the question doesn't state that. And if I did sub in u=0 then I'd get -4.4 = 0 +0a which obviously i can't do because you can't divide by 0

No, velocity is not a linear function of t, so acceleration is not constant so you cant use the (constant acceleration) suvat equations. You have to use calculus so you know acceleration is the rate of change of velocity, so ..

Reply 4

amaniix

OP

3

Original post by mqb2766

No, velocity is not a linear function of t, so acceleration is not constant so you cant use the (constant acceleration) suvat equations. You have to use calculus so you know acceleration is the rate of change of velocity, so ..

Ahh of course. Should have spotted that. So I differentiated t^2 - 0.4t - 4.4 to get 2t-0.4. Then subbed in t=0 to get -0.4m/s^2. Would that be correct?

Reply 5

mqb2766

21

Original post by amaniix

Ahh of course. Should have spotted that. So I differentiated t^2 - 0.4t - 4.4 to get 2t-0.4. Then subbed in t=0 to get -0.4m/s^2. Would that be correct?

Yes and use integration for part of the second part, though for t greater than 4 you can use suvat as the acceleration is constant

(edited 4 months ago)

Reply 6

amaniix

OP

3

Original post by mqb2766

Yes and use integration for the second part.

What steps do I take for b?

Reply 7

mqb2766

21

Original post by amaniix

What steps do I take for b?

for b you have a non suvat phase t<4 and a suvat phase t>4. So youd need to consider each/both and work out what times the particle is 1.2m away from the initial position. Youd need to get the displacement (integrate velocity) from the first phase at t=4, then use that to offset the (suvat) displacement motion for t>4. You could also verify that the velocities are equal at that time point as physically you cant instantaneously jump velocity.

The wording is a bit loose and it could want times when the displacement is -1.2 m as occurs in the first phase (t<4), however the cubic displacement youd be solving is not trivial and its beyond 5 marks. So Id guess they only want solutions in the second phase (t>4) but its not clear from the question. Similarly, Id guess they only want positive displacements so +1.2 m, though there is a negative displacement of -1.2 in the second phase as well.

Cant help thinking this is an "edited" question and theyve not thought too carefully about the changed numbers. There are 5 times when the particle is 1.2m from the initial position, 2 involve solving the cubic and the values are not exact, similarly for the quadratic. More than 5 marks worth of work.

(edited 4 months ago)

Can anyone help with this SUVAT A-level question?

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