A particle has r(0)= (4m)j and v(0)= 92 m/s)i

Question

If tts acceleration is constant and given by a=-(2 m/s^2)(i+j), at what time t does the particle first cross the x axis?
t=2 s, thanks Damon

Also,
at what time t is the partilce moving parallel to the y axis; that is the j direction?

Answer 1

That is when x'(t) = 0

from before: velocity =
r' = [-2 t + x'(0) ] i +[-2 t + y'(0)] j
= [ -2 t +92 ] i - 2 t j

so when -2 t + 92 = 0, there is no velocity component in the i direction
t = 46 seconds

Answer 2

thank you so much!!!

Answer 3

You are welcome.