# A level maths mechanics question

A particle has an initial velocity of (6i - 2j )m s-1 and accelerates for 4 seconds so that its direction of travel is perpendicular to the original direction but its speed is unchanged.
Find the two possible values of the acceleration.?

The answers are meant to be : (-i +2 j ) and (-2i-j) ?
Original post by 1234kelly
A particle has an initial velocity of (6i - 2j )m s-1 and accelerates for 4 seconds so that its direction of travel is perpendicular to the original direction but its speed is unchanged.
Find the two possible values of the acceleration.?

The answers are meant to be : (-i +2 j ) and (-2i-j) ?

Please post your thoughts or working so far. To get you started, perhaps the first thing to consider is the possible direction(s) in which the particle must be travelling after four seconds.
Original post by old_engineer
Please post your thoughts or working so far. To get you started, perhaps the first thing to consider is the possible direction(s) in which the particle must be travelling after four seconds.

I honestly didn’t get anywhere with this question. The particle could travel anywhere in 4 seconds no ? And the part about its speed not changing where do I use that in my solution?
Original post by 1234kelly
I honestly didn’t get anywhere with this question. The particle could travel anywhere in 4 seconds no ? And the part about its speed not changing where do I use that in my solution?

The particle can't travel anywhere. The question states that after four seconds, the direction of travel of the particle is perpendicular to its original direction of travel (which is given) and that its speed is unchanged. So after four seconds the velocity of the particle will be, say, (ai + bj). The constraints on (ai + bj) are that (ai + bj) is perpendicular to (6i - 2j) and the magnitude of (ai + bj) is the same as the magnitude of (6i - 2j).
Original post by old_engineer
The particle can't travel anywhere. The question states that after four seconds, the direction of travel of the particle is perpendicular to its original direction of travel (which is given) and that its speed is unchanged. So after four seconds the velocity of the particle will be, say, (ai + bj). The constraints on (ai + bj) are that (ai + bj) is perpendicular to (6i - 2j) and the magnitude of (ai + bj) is the same as the magnitude of (6i - 2j).

So the possible directions are (-6i -2j ) and (6i + 2j) ? Wait are those possible directions the values of my final velocity and so then I just then use SUVAT with u = (6i -2j) , V= (-6i -2j ) and (6i + 2j ) T=4 to work out the 2 acceleration? But I did that and I definitely dont get the right answers I really don’t think in the right direction could you please explain what I’m just not seeing I still don’t get how to get to get the final answer .
(edited 6 months ago)
Original post by 1234kelly
So the possible directions are (-6i -2j ) and (6i + 2j) ? Wait are those possible directions the values of my final velocity and so then I just then use SUVAT with u = (6i -2j) , V= (-6i -2j ) and (6i + 2j ) T=4 to work out the 2 acceleration? But I did that and I definitely dont get the right answers I really don’t think in the right direction could you please explain what I’m just not seeing I still don’t get how to get to get the final answer .

(-6i-2j) and (6i+2j) are not perpendicular to 6i - 2j (e.g. (6i+2j).(6i - 2j) = 36 - 4 = 32).

(The rest of what you suggest seems reasonable).
(edited 6 months ago)
Original post by DFranklin
(-6i-2j) and (6i+2j) are not perpendicular to 6i - 2j (e.g. (6i+2j).(6i - 2j) = 36 - 4 = 32).

(The rest of what you suggest seems reasonable).

Wait so how do I work out the the perpendicular vector? I originally just sketched it and came up with those 2 answers.
Original post by 1234kelly
Wait so how do I work out the the perpendicular vector? I originally just sketched it and came up with those 2 answers.

Two vectors perpendicular to $(a{\bf i} + b{\bf j})$ are $\pm (b{\bf i} - a{\bf j})$
suppose the original velocity vector is pi + qj

the new, perpendicular vector is either qi - pj ( rotation of 90 degrees clockwise )

or the opposite which is -qi + pj ( rotation of 90 degrees anticlockwise )