Question

A particle moves along a path y=x^2+4x+100, starting with an initial velocity of Vo=(4i-16j) m/s. If Vx is constant, what is value of Vy and ay at x=16m?

Answer 1

Explanation:

V= dy/dt = 2x+4

x=16m

2*16+4= 36

Vy = 36*4= 144m/s

here 4 is the value of velocity in x direction.