Question

A particle is projected with a velocity $\vec{v} = (2\hat{i} + 3\hat{j})$ m/s in the presence of a uniform acceleration $\vec{a} = (-3\hat{i} - 4\hat{j})$ m/s². The path of the particle will be

• A. Straight line
• B. Parabolic
• C. Circular
• D. Elliptical

Answer 1

Answer: (B)

Parabolic

Answer 2

The position vector of a particle at time $t$ moving with initial velocity $\vec{v}_0$ and uniform acceleration $\vec{a}$ is given by:

$\vec{r}(t) = \vec{r}(0) + \vec{v}_0 t + \frac{1}{2}\vec{a}t^2$

Let the initial position be the origin, $\vec{r}(0) = \vec{0}$. Given initial velocity $\vec{v}_0 = (2\hat{i} + 3\hat{j})$ m/s and uniform acceleration $\vec{a} = (-3\hat{i} - 4\hat{j})$ m/s². The position vector at time $t$ is:

$\vec{r}(t) = (2\hat{i} + 3\hat{j})t + \frac{1}{2}(-3\hat{i} - 4\hat{j})t^2$ $\vec{r}(t) = (2t - \frac{3}{2}t^2)\hat{i} + (3t - 2t^2)\hat{j}$

Let the coordinates of the particle at time $t$ be $(x(t), y(t))$.

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

To find the path, we eliminate $t$ from these two equations. Multiply the first equation by 4 and the second by 3:

$4x = 8t - 6t^2$ $3y = 9t - 6t^2$

Subtract the first equation from the second:

$3y - 4x = (9t - 6t^2) - (8t - 6t^2)$ $3y - 4x = t$

Now substitute this expression for $t$ back into either of the original equations. Using the equation for $y$:

$y = 3(3y - 4x) - 2(3y - 4x)^2$ $y = 9y - 12x - 2(9y^2 - 24xy + 16x^2)$ $y = 9y - 12x - 18y^2 + 48xy - 32x^2$

Rearranging the terms to the left side:

$32x^2 + 18y^2 - 48xy + 12x + y - 9y = 0$ $32x^2 + 18y^2 - 48xy + 12x - 8y = 0$

This is a general second-degree equation in $x$ and $y$ of the form $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$. Here, $A = 32$, $B = -48$, $C = 18$. The nature of the conic section is determined by the discriminant $B^2 - 4AC$.

$B^2 - 4AC = (-48)^2 - 4(32)(18) = 2304 - 4(576) = 2304 - 2304 = 0$.

Since $B^2 - 4AC = 0$, the path is a parabola (or a degenerate case like parallel lines). In the case of motion under constant acceleration where the initial velocity is not parallel to the acceleration, the path is a parabola.

We can check if the initial velocity $\vec{v}_0 = (2, 3)$ and acceleration $\vec{a} = (-3, -4)$ are parallel. Two vectors $\vec{u}$ and $\vec{v}$ are parallel if $\vec{u} = k\vec{v}$ for some scalar $k$. Is $(2, 3) = k(-3, -4)$?

$2 = -3k \implies k = -2/3$ $3 = -4k \implies k = -3/4$

Since the value of $k$ is not unique, the vectors are not parallel.

Therefore, the path is a parabola.