A Level Maths Mechanics

(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

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

(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 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.

(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