"A particle moves in the horizontal plane that contains the perpendicular unit vectors i and j. Initially it is at the origin and has velocity 18ims^-1. After accelerating for 10 seconds its velocity is (30i + 8j)ms^-1. Assume that the acceleration of the particle is constant.
a) Find the acceleration of the particle
b) Find the position vector of the particle when the velocity is (36i + 12j)ms^-1.
2. The attempt at a solution
For (a) i managed to find out the acceleration being (1.2i + 0.8j)ms^-2 using v=at+u and rearranging it to a=(v-u)/t.
But for (b) it says the answer is 405i + 90j
I've tried it over and over again but I don't think the question was specific enough (probably just me not understanding it).
I've tried it this way using:
u=18i + 0j
v=36i + 12j
R (position vector; i used R=(t/2)(u+v))=5(u+v) which gives the vector R= 240i + 40j
I've tried it by replacing the initial velocity with 36i + 12j but that didn't work either. I can probably work out the question if I understood what it basically is asking for.
Someone Please run me through this, I'm not good at mechanics :'(