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A Level Maths Question (1)

A particle P is initially at the point (2,6) in relation to an origin O. It is moving with velocity (3i + j) and constant acceleration (16i + 24j) , show that in 2 seconds, it is moving directly away from O.

- for this question my initial thought was just to find the final velocity and then prove that V is a multiple of U i.e its parallel and since it starts from the same point its moved in the same direction. But V= 35i +49J so my proof wont work.
1) what's wrong with the logic I've used ? a particle has to travel in the same direction as it's velocity no ?
2) what's a better/ correct solution ?
Reply 1
Original post by 1234kelly
A particle P is initially at the point (2,6) in relation to an origin O. It is moving with velocity (3i + j) and constant acceleration (16i + 24j) , show that in 2 seconds, it is moving directly away from O.

- for this question my initial thought was just to find the final velocity and then prove that V is a multiple of U i.e its parallel and since it starts from the same point its moved in the same direction. But V= 35i +49J so my proof wont work.
1) what's wrong with the logic I've used ? a particle has to travel in the same direction as it's velocity no ?
2) what's a better/ correct solution ?

If its moving directly away from the origin at a point in time, then the current velocity v must be a positive scalar multiple of the position.

A sketch of the postiion (as a point) and the velocity as an arrow with the tail at the point/current position should make it clear?
(edited 3 months ago)
Reply 2
Original post by mqb2766
If its moving directly away from the origin at a point in time, then the current velocity v must be a positive scalar multiple of the position.

A sketch of the postiion (as a point) and the velocity as an arrow with the tail at the point/current position should make it clear?

oh that makes sense thank you! but why isn't the velocity being a positive scalar multiple of the initial velocity proof for it travelling in the same direction is that wrong to assume ?
Reply 3
Original post by 1234kelly
oh that makes sense thank you! but why isn't the velocity being a positive scalar multiple of the initial velocity proof for it travelling in the same direction is that wrong to assume ?

You could "calculate" the positions and velocities at t=0 and t=2 and sketch them. There isnt really a relationship between them in the way you seem to be thinking. You integrate acceleration (twice) to get the position curve which joins the two points (and beyond for t>2) and you imagine the current velocity arrow (at time t) as being tangential to the positiion curve at time (t) as in the previous post.

So something like
https://www.statisticshowto.com/tangent-vector-velocity/
where T is the velocity vector at each point (position/time). The first point on the curve (assuming it starts at O) is something like this question is after.
(edited 3 months ago)
Reply 4
Original post by mqb2766
You could "calculate" the positions and velocities at t=0 and t=2 and sketch them. There isnt really a relationship between them in the way you seem to be thinking. You integrate acceleration (twice) to get the position curve which joins the two points (and beyond for t>2) and you imagine the current velocity arrow (at time t) as being tangential to the positiion curve at time (t) as in the previous post.

So something like
https://www.statisticshowto.com/tangent-vector-velocity/
where T is the velocity vector at each point (position/time). The first point on the curve (assuming it starts at O) is something like this question is after.

Thank you for attaching the link it definitely help !
Reply 5
Original post by mqb2766
If its moving directly away from the origin at a point in time, then the current velocity v must be a positive scalar multiple of the position.

A sketch of the postiion (as a point) and the velocity as an arrow with the tail at the point/current position should make it clear?

Sorry to bother you again with this but what would it mean if the velocity wasn’t multiple of displacement? Is it possible? Would it it just not pass through the origin ?
Reply 6
Original post by 1234kelly
Sorry to bother you again with this but what would it mean if the velocity wasn’t multiple of displacement? Is it possible? Would it it just not pass through the origin ?

The velocity is not usually (rarely) a multiple of the position. Just draw a/any position curve and and for any point (position) on the curve, the velocity vector points in the same direction as the tangent (by definition). Tracing the velocity vector/tangent backwards, its rare that it will pass through the origin as the previous link illustrated, assuming the origin is defined at the start of the curve in the bottom left.

You imagine the current position as a point in space and the velocity is a vector with a tail at that point. Obviously, it could point in any direction and only one of those would correspond to it tracing backwards through the origin.

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