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A level maths mechanics question (1)

For this question I understand the markscheme but for part b when finding the displacement how is it possible that we can use both integration and SUVAT? I thought integration was only for variable acceleration and suvat had to be only for constant?8BEF39B8-D59A-416A-8A9E-22DF517C43C3.jpeg
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(edited 1 month ago)
Reply 1
Suvat is when you integrate (differentiate) the constant acceleration case, so either use the formulae directly or (long way) integrate a constant acceleration so
a <-> u + at <-> ut + 1/2 at^2
where you integrate/differentiate wrt t
(edited 1 month ago)
Reply 2
Original post by mqb2766
Suvat is when you integrate (differentiate) the constant acceleration case, so either use the formulae directly or (long way) integrate a constant acceleration so
a <-> u + at <-> ut + 1/2 at^2
where you integrate/differentiate wrt t

-So for constant acceleration you can use both integration/ differentiation and suvat its just quicker and preferred to use SUVAT ?

-but for variable acceleration questions you can only use integration/ differentiation ?
Reply 3
Original post by 1234kelly
-So for constant acceleration you can use both integration/ differentiation and suvat its just quicker and preferred to use SUVAT ?
-but for variable acceleration questions you can only use integration/ differentiation ?

Yes, suvat is simply the constant acceleration case where youve done the (definite) integral and the u,v are the velocity limits and 0,s are the displacement limits and 0,t the time limits.

For "interest", the time integration/differentiation case is straight forward so
acceleration <-> velocity <-> displacement
as in the previous post. The
v - u = at
suvat is simply change in momentum = impulse as multiplying through by m
mv - mu = (ma)t
so an applied force=ma over a time t (impulse) will produce a change in momentum mv-mu.

Similarly, the
v^2 - u^2 = 2as
can represent the conservation of energy as multiplying through by m/2 gives
(1/2)mv^2 - (1/2)mu^2 = mas
so a=g and s=h and you have
change in KE = change in GPE. You can also derive it from the constant acceleration assumption as
a = dv/dt = dv/ds ds/dt = v dv/ds
and integrate (definite) both sides with respect to s you get
as = (1/2) v^2 - (1/2) u^2

So the basic suvat equations can be relatively interesting if you make the links.
(edited 1 month ago)
Reply 4
Original post by mqb2766
Yes, suvat is simply the constant acceleration case where youve done the (definite) integral and the u,v are the velocity limits and 0,s are the displacement limits and 0,t the time limits.
For "interest", the time integration/differentiation case is straight forward so
acceleration <-> velocity <-> displacement
as in the previous post. The
v - u = at
suvat is simply change in momentum = impulse as multiplying through by m
mv - mu = (ma)t
so an applied force=ma over a time t (impulse) will produce a change in momentum mv-mu.
Similarly, the
v^2 - u^2 = 2as
can represent the conservation of energy as multiplying through by m/2 gives
(1/2)mv^2 - (1/2)mu^2 = mas
so a=g and s=h and you have
change in KE = change in GPE. You can also derive it from the constant acceleration assumption as
a = dv/dt = dv/ds ds/dt = v dv/ds
and integrate (definite) both sides with respect to s you get
as = (1/2) v^2 - (1/2) u^2
So the basic suvat equations can be relatively interesting if you make the links.

Omg thank you that’s genuinely helped a lot and your right it is interesting especially as I take physics

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