A particle starts from the origin at t = 0 with an
initial velocity of 3.0 î m/s and moves in the
x-y plane with a constant acceleration ( 6î + 4ĵ ) m/s² . The x-coordinate of the particle at the instant when its y-coordinate is 32 m is D meters. The value of D is :-

Solution

Physics sir · 3 steps

Treat x and y separately: the y-motion gives you the time, then put that time into x.

1. Split the motion. Along y the particle starts with zero velocity and has a_y = 4 m s⁻². Along x it starts with uₓ = 3 m/s and has aₓ = 6 m s⁻².
2. Find the time. y = 1/2a_y t² = 2t². Setting 2t² = 32 gives t² = 16, so t = 4 s.

   x = uₓ t + 1/2aₓ t² = 3(4) + 1/2(6)(4)²

3. So. x = 12 + 48 = 60 m, so D = 60, option (d).

The idea to keep. With constant acceleration in 2D, each axis follows its own s = ut + 1/2at². The only thing the two axes share is time, so use one axis to find t.

Answer from the NTA final answer key. Solution by PrepWiser.