The co-ordinates of a particle moving in x-y plane are given by : x=2+4 t, y=3 t+8 t².
The motion of the particle is :

Solution

Physics sir · 4 steps

Differentiate twice for the acceleration, then check whether the velocity starts along it.

1. Velocity. Differentiate each coordinate once: vₓ = dx/dt = 4 and v_y = dy/dt = 3 + 16t. At t = 0 the velocity is (4, 3).
2. Acceleration. Differentiate again: aₓ = 0 and a_y = 16. The acceleration is a constant 16 along +y, so the motion is uniformly accelerated.
3. Shape of path. The initial velocity (4, 3) is not along the acceleration (0, 16), so the path bends. Remove t using t = (x-2)/4 to see the shape.

   y = 3/4(x-2) + 1/2(x-2)²

4. So. y is quadratic in x, so the path is a parabola. The motion is uniformly accelerated along a parabolic path, option (d).

The idea to keep. Constant acceleration gives a straight line only if the initial velocity points along the acceleration. If they are at an angle, the path is a parabola, just like a projectile.

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