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as level mechanics suvat

Question:
The point A is 1.8 m vertically above horizontal ground.
At time t = 0, a small stone is projected vertically upwards with speed U m s−1
from the
point A.
At time t = T seconds, the stone hits the ground.
The speed of the stone as it hits the ground is 10 m s−1
In an initial model of the motion of the stone as it moves from A to where it hits
the ground
the stone is modelled as a particle moving freely under gravity
the acceleration due to gravity is modelled as having magnitude 10 m s−2
Using the model,
(a) find the value of U

suvat answer:
s = -1.8
u = U
v = -10
a = -10
t = ?

I understand why you use these values and that upwards is the positive direction but I dont get why the displacement is negative and the final velocity is negative. Is the answer based on the whole upwards and downwards motion of the stone or is it based on the upwards/downwards motion of the stone only?
Reply 1
Original post by ljaiju
Question:
The point A is 1.8 m vertically above horizontal ground.
At time t = 0, a small stone is projected vertically upwards with speed U m s−1
from the
point A.
At time t = T seconds, the stone hits the ground.
The speed of the stone as it hits the ground is 10 m s−1
In an initial model of the motion of the stone as it moves from A to where it hits
the ground
the stone is modelled as a particle moving freely under gravity
the acceleration due to gravity is modelled as having magnitude 10 m s−2
Using the model,
(a) find the value of U
suvat answer:
s = -1.8
u = U
v = -10
a = -10
t = ?
I understand why you use these values and that upwards is the positive direction but I dont get why the displacement is negative and the final velocity is negative. Is the answer based on the whole upwards and downwards motion of the stone or is it based on the upwards/downwards motion of the stone only?

At A, displacement 0 occurs at t=0, so the final displacement is 1.8m below that position so s=-1.8 as its negative downwards. The final velocity is also heading downwards so v=-10.

It may help to sketch the parabola of vertical displacement against time. If youre unsure, post what you think.
(edited 10 months ago)
Reply 2
Original post by mqb2766
displacement 0 occurs at t=0, so the final displacement is 1.8m below that position so s=-1.8 as its negative downwards. The final velocity is also heading downwards so v=-10.
It may help to sketch the parabola of vertical displacement against time. If youre unsure, post what you think.

that is fine, i understand it much better now thank youu.
Reply 3
Original post by mqb2766
At A, displacement 0 occurs at t=0, so the final displacement is 1.8m below that position so s=-1.8 as its negative downwards. The final velocity is also heading downwards so v=-10.
It may help to sketch the parabola of vertical displacement against time. If youre unsure, post what you think.

Hii im having trouble with the same question. I understand why the displacement and the velocity is negative but if the stone is coming back down to the ground wouldn't the acceleration be positive because gravity is acting normally?
Reply 4
Original post by fa.17
Hii im having trouble with the same question. I understand why the displacement and the velocity is negative but if the stone is coming back down to the ground wouldn't the acceleration be positive because gravity is acting normally?

You define the positive direction and in the OP, positive is defined as upwards. It could be defined as downwards and as long as the maths is done consistently in both cases, the answer will work out as physics (gravity) doesnt care which way you think is positive.

So if its positive up, then the final velocity and displacement are negative. Acceleration is also negative as gravity acts downwards. If the acceleration was positive, the stone would accelerate off into space. If youre happy that velocity is reducing and goes from positive to negative, then acceleration is the rate of change of velocity, so drawing a velocity time graph where the (linear) velocity goes from positive to negative, then the gradient or acceleration is negative so -g or -9.8.

For this question part (dont know what else there is) it would make sense to define positive as down as u is the only thing that points up, but s,v, a are all downwards and its generally simpler to choose the positive direction to make the most suvat terms positive as it simplfiies the calculation.
(edited 8 months ago)

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