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Vectory, velocity kinematics. HELP

A particle P moves with constant acceleration (i–2j) m s–2.At time t= 0, the particle is at the point A with position vector (2i+ 5j) m and is moving with velocity um s–1.At time t= 3 s, Pis at the point B with position vector (–2.5i+ 8j) m.
Find u.

Any help would be appreciated :smile:
Well, what have you tried?
It would be helpful to see which method you tried, because I have two in mind...
Reply 2
Original post by tonyiptony
Well, what have you tried?
It would be helpful to see which method you tried, because I have two in mind...

I have subbed the values 2i +5j, -2.5i +8j and 3 in the formulae r = r 0 + v t to get 2i +5j = -2.5i +8j + 3v ans then rearranged to get 1.5i-j = v. Then subbed these values into thr formulae v = u + at to get 1.5i -j = u + (i-2j)3 then got the value of u to be -1.5 + 5j. However the answer is (-3i + 4)
(I am new to this topic so apologise if I've broken maths here😅)
Original post by LoneWonder
I have subbed the values 2i +5j, -2.5i +8j and 3 in the formulae r = r 0 + v t to get 2i +5j = -2.5i +8j + 3v ans then rearranged to get 1.5i-j = v. Then subbed these values into thr formulae v = u + at to get 1.5i -j = u + (i-2j)3 then got the value of u to be -1.5 + 5j. However the answer is (-3i + 4)
(I am new to this topic so apologise if I've broken maths here😅)

Ah, okay, you've gone the suvat route.
I think what you've gone wrong is using r=r0+utr = r_{0} + ut - this formula assumes the particle is moving at constant velocity and with no acceleration. (Beware of the "v" here. This v is almost but not quite the the same v in suvat.)

But the idea is there - you just need to factor in acceleration... but wait, we already have a suvat formula for it, namely s=ut+12at2\displaystyle s=ut+\frac{1}{2}at^{2}. (In fact, r=r0+utr = r_{0} + ut is just a simplified version of this. We set a=0 and s to actually calculate displacement, and we'll get that.)

Of course the problem is the s. This s is the displacement, not the position. So if you know what displacement is (as in if you really know the concept of it), you should be good to go.
(edited 5 months ago)
Reply 4
Original post by tonyiptony
Ah, okay, you've gone the suvat route.
I think what you've gone wrong is using r=r0+utr = r_{0} + ut - this formula assumes the particle is moving at constant velocity and with no acceleration. (Beware of the "v" here. This v is almost but not quite the the same v in suvat.)

But the idea is there - you just need to factor in acceleration... but wait, we already have a suvat formula for it, namely s=ut+12at2\displaystyle s=ut+\frac{1}{2}at^{2}. (In fact, r=r0+utr = r_{0} + ut is just a simplified version of this. We set a=0 and s to actually calculate displacement, and we'll get that.)

Of course the problem is the s. This s is the displacement, not the position. So if you know what displacement is (as in if you really know the concept of it), you should be good to go.

Managed to get the answer thanks:smile:

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