# A Level Maths Mechanics

Watch
Announcements

Page 1 of 1

Go to first unread

Skip to page:

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

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

0

reply

Report

#2

(Original post by

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

**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

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

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

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

In fact, you can write down the displacement vector using the equation where is the position vector A, is the velocity vector at A, and is the acceleration.. with 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 (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 component equal to component, and they must be positive.

Since the position of P is described by we can use this fact and equation the two components, hence obtain a quadratic for and solve it.

Last edited by RDKGames; 2 years ago

0

reply

(Original post by

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 where is the position vector A, is the velocity vector at A, and is the acceleration.. with 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 (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 component equal to component, and they must be positive.

Since the position of P is described by we can use this fact and equation the two components, hence obtain a quadratic for and solve it.

**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 where is the position vector A, is the velocity vector at A, and is the acceleration.. with 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 (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 component equal to component, and they must be positive.

Since the position of P is described by we can use this fact and equation the two components, hence obtain a quadratic for and solve it.

0

reply

Report

#4

(Original post by

so for part a I subtracted (7i-10j) from (2i-3j) divising it all by 2 as that is time taken to get acceleration

**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

**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 formula on it instead (just as you would in part b). You know s, u, and t. Solve for a.

Last edited by RDKGames; 2 years ago

0

reply

Report

#5

(Original post by

That's incorrect, because 7i-10j is a

EDIT: And TBH, I think you may as well just slap on the formula on it instead (just as you would in part b). You know s, u, and t. Solve for a.

**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 formula on it instead (just as you would in part b). You know s, u, and t. Solve for a.

0

reply

X

Page 1 of 1

Go to first unread

Skip to page:

### Quick Reply

Back

to top

to top