Question

A particle is moving in x-y plane such that x=(4t^2+1)m,y=(8t+2)m.the initial velocity of particle will be?

Answer 1

dx/dt=8t,Dy/dtt=8
initially at t=0.
v along x=0, along y=8
therefore velovity=8j

Answer 2

x = 4t² + 1
v_x = dx/dt = 8t
At t = 0
v_x = 8 × 0 = 0

y = 8t + 2
v_y = dy/dt = 8
At t = 0
v_y = 8 m/s

So,
Initial velocity is 8 m/s along y-axis
If j is the unit vector along y-axis then initial velocity is 8j m/s