A mass is moving with constant acceleration in the x-y plane. At time t = 0s the mass’s position is 2mˆı+3mjˆ, the mass’s velocity is 3 m/s ˆı −4 m / s jˆ and the mass is subject to an acceleration of 1 m/ s ^2 ˆı + 1 m/ s^ 2 i jˆ.
At what time is the velocity perpendicular to the acceleration?
v = v0 + α t
x = x0 + v0t + 0.5at^2
v^2= v0^2 +2a(x-x0)
The Attempt at a Solution:
I don't really know how to approach this question. I tried to take the magnitude of velocity and acceleration and find the radius using a= v^2/R, but I don't think that is the way to solve it.
I know that the answer is t=0.5 sec.
Many thanks for your help