cheese13
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Can I please get some help on these 2 part?

In this question i and j are unit vectors

A particle P moves with constant acceleration
At time t=0, the particle is at O and is moving with velocity (2i-3j)ms^-2
At time t=2 seconds, P is at the point A with positive vector (7i-10j)m

Show that the magnitude of the acceleration of P is 2.5ms^-2

At that instant when P leaves the point A, the acceleration of P changes so that P moves with constant acceleration (4i+8.8j)ms^-2

At the instant when P reaches the point B, the direction of P is north east

Find the time it takes for P to travel from A to B

Thank You
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RDKGames
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(Original post by aimman_s)
Can I please get some help on these 2 part?

In this question i and j are unit vectors

A particle P moves with constant acceleration
At time t=0, the particle is at O and is moving with velocity (2i-3j)ms^-2
At time t=2 seconds, P is at the point A with positive vector (7i-10j)m
Use these hints

Show that the magnitude of the acceleration of P is 2.5ms^-2
Since acceleration is constant, you can just determine the difference in velocities and divide by time taken to get the acceleration vector.
So consider its magnitude for the answer. So as a first step, try to figure out what the velocity must be at A.

At that instant when P leaves the point A, the acceleration of P changes so that P moves with constant acceleration (4i+8.8j)ms^-2

At the instant when P reaches the point B, the direction of P is north east

Find the time it takes for P to travel from A to B

Thank You
So the particle now has a different acceleration vector, which is still constant.

In fact, you can write down the displacement vector using the equation \mathbf{s} =\mathbf{s}_0 + \mathbf{u} t + \dfrac{1}{2} \mathbf{a} t^2 where \mathbf{s}_0 is the position vector A, \mathbf{u} is the velocity vector at A, and \mathbf{a} is the acceleration.. with t being the time, of course.

You are told that when P reaches B, it is in the north-east direction. So now think about what this means vector wise. This must mean that it's on the vector \mathbf{i} + \mathbf{j} (or some arbitrary size of it... it's just the direction that we care about) but the point is that every single position which we would consider to be 'north-east' must have it's \mathbf{i} component equal to \mathbf{j} component, and they must be positive.

Since the position of P is described by \mathbf{s} we can use this fact and equation the two components, hence obtain a quadratic for t and solve it.
Last edited by RDKGames; 2 years ago
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cheese13
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(Original post by RDKGames)
Use these hints



Since acceleration is constant, you can just determine the difference in velocities and divide by time taken to get the acceleration vector.
So consider its magnitude for the answer. So as a first step, try to figure out what the velocity must be at A.



So the particle now has a different acceleration vector, which is still constant.

In fact, you can write down the displacement vector using the equation \mathbf{s} =\mathbf{s}_0 + \mathbf{u} t + \dfrac{1}{2} \mathbf{a} t^2 where \mathbf{s}_0 is the position vector A, \mathbf{u} is the velocity vector at A, and \mathbf{a} is the acceleration.. with t being the time, of course.

You are told that when P reaches B, it is in the north-east direction. So now think about what this means vector wise. This must mean that it's on the vector \mathbf{i} + \mathbf{j} (or some arbitrary size of it... it's just the direction that we care about) but the point is that every single position which we would consider to be 'north-east' must have it's \mathbf{i} component equal to \mathbf{j} component, and they must be positive.

Since the position of P is described by \mathbf{s} we can use this fact and equation the two components, hence obtain a quadratic for t and solve it.
so for part a I subtracted (7i-10j) from (2i-3j) divising it all by 2 as that is time taken to get acceleration
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RDKGames
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(Original post by aimman_s)
so for part a I subtracted (7i-10j) from (2i-3j) divising it all by 2 as that is time taken to get acceleration
That's incorrect, because 7i-10j is a position vector, whereas 2i-3j is a velocity vector. You cannot subtract them and get a meaningful quantity.

EDIT: And TBH, I think you may as well just slap on the s=ut+\frac{1}{2}at^2 formula on it instead (just as you would in part b). You know s, u, and t. Solve for a.
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JW8680
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(Original post by RDKGames)
That's incorrect, because 7i-10j is a position vector, whereas 2i-3j is a velocity vector. You cannot subtract them and get a meaningful quantity.

EDIT: And TBH, I think you may as well just slap on the s=ut+\frac{1}{2}at^2 formula on it instead (just as you would in part b). You know s, u, and t. Solve for a.
Just run into this question myself, managed to do part a but a little stuck with b, any help would be appreciated
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