help 27

The question states that the particle     'is moving in uniform acceleration'       It does NOT say that x and y are coordinates, they are distances.

Let's look at the position x   x= vot + 1/2 at^2       vo = original velocity

solve for vo    

vo = x/t- 1/2at

Now, the VELOCITY at x    Vx= vo + at

Substitute for vo  to get 

Vx = (x/t -1/2at) +at

Vx = x/t +1/2at

Now, for y 

y =vt' + 1/2at'2      In THIS instance  the 'v' is Vx which we found above, so let's substitute this into the equation

y= (x/t + 1/2at) t'  + 1/2at'^2   now solve for a

y-xt'/t = 1/2a (tt' +t'2)

2 (y-xt'/t) = a t'(t+t')

a= 2(y-xt'/t) / (t' (t+t') )           Simplify

a= 2(y/t' -x/t) / (t+t')

Sorry EP but that is not correct.

It's only correct if the acceleration in the x direction is the same as the acceleration in the y direction, and they need not be.