Question

A particle starts from the origin at t = 0s with a velocity of \(10\hat{j}\) ms^-1 and move in the x-y plane with a constant acceleration of \((8\hat{i}+2\hat{j})\) ms^-2 . At an instant when the x-coordinate of the particle is 16 m, y-coordinate of the particle is:

• A. 16 m
• B. 28 m
• C. 36 m
• D. 24 m

Answer 1

The Correct Option is D

1. Analyze the given information:

• Initial position: Origin (0, 0)
• Initial velocity: $10\hat{j}$ ms^-1
• Acceleration: $(8\hat{i} + 2\hat{j})$ ms^-2
• x-coordinate at a certain instant: 16 m

2. Find the time:

We can use the equation of motion for the x-direction:

$x = u_xt + \frac{1}{2}a_xt^2$

Since the initial velocity is in the y-direction, $u_x = 0$. The x-component of acceleration is $a_x = 8$ m/s². We are given $x = 16$ m. Substituting these values:

$16 = 0 \cdot t + \frac{1}{2} \cdot 8 \cdot t^2$

$16 = 4t^2$

$t^2 = 4$

$t = 2$ s (We take the positive value of time)

3. Find the y-coordinate:

Now we can use the equation of motion for the y-direction:

$y = u_yt + \frac{1}{2}a_yt^2$

The initial velocity in the y-direction is $u_y = 10$ m/s, the y-component of acceleration is $a_y = 2$ m/s², and we found $t = 2$ s. Substituting these values:

$y = 10 \cdot 2 + \frac{1}{2} \cdot 2 \cdot 2^2$

$y = 20 + 4$

$y = 24$ m

The correct answer is (D) 24 m.

Answer 2

1. Understanding the Kinematic Equations: 
We'll use the following kinematic equation for constant acceleration: $$ \mathbf{r} = \mathbf{r}_0 + \mathbf{v}_0 t + \frac{1}{2} \mathbf{a} t^2 $$ Where:

• $\mathbf{r}$ is the final position vector,
• $\mathbf{r}_0$ is the initial position vector (0, 0),
• $\mathbf{v}_0$ is the initial velocity vector,
• $\mathbf{a}$ is the acceleration vector,
• $t$ is the time.

2. Given Values:

• $\mathbf{r}_0 = 0 \mathbf{i} + 0 \mathbf{j}$,
• $\mathbf{v}_0 = 0 \mathbf{i} + 10 \mathbf{j} \, \text{m/s}$,
• $\mathbf{a} = 8 \mathbf{i} + 2 \mathbf{j} \, \text{m/s}^2$,
• $x = 16 \, \text{m}$.

3. Breaking Down the Equation into Components:

• For the x-component: $$ x = x_0 + v_{0x} t + \frac{1}{2} a_x t^2 $$
• For the y-component: $$ y = y_0 + v_{0y} t + \frac{1}{2} a_y t^2 $$

4. Solving for Time ($t$) Using the x-Component:

$$ 16 = 0 + 0 \cdot t + \frac{1}{2} \cdot 8 \cdot t^2 $$ $$ 16 = 4t^2 $$ $$ t^2 = 4 $$ $$ t = 2 \, \text{s} \quad (\text{We take the positive root for time.}) $$

5. Solving for the y-Coordinate:

$$ y = 0 + 10 \cdot t + \frac{1}{2} \cdot 2 \cdot t^2 $$ Substitute $t = 2$: $$ y = 10 \cdot 2 + \frac{1}{2} \cdot 2 \cdot (2)^2 $$ $$ y = 20 + 4 $$ $$ y = 24 \, \text{m} $$

Final Answer: The correct answer is (D) 24 m.