The co-ordinates of a particle moving in x-y plane are given by: x = 2 + 4t, y = 3t + 8t². The motion of the particle is: (a) uniform motion along a straight line. (b) non-uniformly accelerated. (c) uniformly accelerated having motion along a straight line. (d) uniformly accelerated having motion along a parabolic path.

Answer: (d). vx = 4 (constant), vy = 3 + 16t, so a = 16 j (constant): uniformly accelerated Initial velocity 4 i + 3 j is not along a, so the path is curved Eliminating t = x-2/4 gives y quadratic in x: a parabola So the answer is (d).

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Motion in a Plane (2-D Kinematics)

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

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Answer: (d)

$v_x = 4$ (constant), $v_y = 3 + 16t$ , so $\vec a = 16\hat j$ (constant): uniformly accelerated
Initial velocity $4\hat i + 3\hat j$ is not along $\vec a$ , so the path is curved
Eliminating $t = \frac{x-2}{4}$ gives $y$ quadratic in $x$ : a parabola
So the answer is (d).

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