Given position coordinates

$x=2+4t, \qquad y=3t+8t^2$

So the position vector is

$\vec r = x\hat i + y\hat j = (2+4t)\hat i + (3t+8t^2)\hat j$

Find velocity components

Velocity is the time derivative of position:

$v_x=\frac{dx}{dt}=4, \qquad v_y=\frac{dy}{dt}=3+16t$

Hence,

$\vec v = 4\hat i + (3+16t)\hat j$

Find acceleration components

Acceleration is the time derivative of velocity:

$a_x=\frac{dv_x}{dt}=0, \qquad a_y=\frac{dv_y}{dt}=16$

Hence,

$\vec a = 0\hat i + 16\hat j$

Since acceleration is constant, the motion is uniformly accelerated.

Find the path of the particle

From

$x=2+4t$

we get

$t=\frac{x-2}{4}$

Substitute into the equation for $y$:

$y=3\left(\frac{x-2}{4}\right)+8\left(\frac{x-2}{4}\right)^2$

Simplify:

$y=\frac{3(x-2)}{4}+8\cdot \frac{(x-2)^2}{16}$

$y=\frac{3(x-2)}{4}+\frac{(x-2)^2}{2}$

This is a quadratic equation in $x$, so the trajectory is a parabola.

Check options

• A: Uniform motion along a straight line.
  False, because acceleration is not zero and path is not straight.

• B: Non-uniformly accelerated.
  False, because acceleration is constant.

• C: Uniformly accelerated having motion along a straight line.
  False, because the path is parabolic, not straight.

• D: Uniformly accelerated having motion along a parabolic path.
  True.

Final answer

The particle has constant acceleration and moves along a parabolic path.