Question

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

• A. non-uniformly accelerated.
• B. uniformly accelerated having motion along a straight line.
• C. uniform motion along a straight line.
• D. uniformly accelerated having motion along a parabolic path.

Answer 1

Answer: (D)

uniformly accelerated having motion along a parabolic path.

Answer 2

Calculate the Velocity Components:

For $x = 2 + 4t$:

$\frac{dx}{dt} = v_x = 4$

For $y = 3t + 8t^2$:

$\frac{dy}{dt} = v_y = 3 + 16t$

Calculate the Acceleration Components:

The acceleration in the $x$-direction $a_x$ is:

$\frac{d^2x}{dt^2} = a_x = 0$

The acceleration in the $y$-direction $a_y$ is:

$\frac{d^2y}{dt^2} = a_y = 16$

Therefore, the particle has a constant acceleration $a_y = 16\, \text{m/s}^2$ in the $y$-direction, and no acceleration in the $x$-direction.

Determine the Path of Motion:

To find the path of the particle, express $y$ in terms of $x$ by eliminating $t$ between the two equations.

From $x = 2 + 4t$, we get:

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

Substitute this into the equation for $y$:

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

Simplifying, we get:

$y = \frac{3}{4}(x - 2) + 8 \times \frac{(x - 2)^2}{16}$ $y = \frac{3}{4}(x - 2) + \frac{1}{2}(x - 2)^2$

This equation is quadratic in $x$, indicating that the path of the particle is a parabola.

Conclusion:

The motion of the particle is uniformly accelerated with a parabolic path, which corresponds to Option (4).