JEE Main 2024 — 27 January, Shift 1 — Question 48

A particle starts from the origin at $t=0$, so its initial position vector is $\vec{r}_0 = 0\hat{i} + 0\hat{j}$. The initial velocity is given as $\vec{v}_0 = 5\hat{i} \text{ m/s}$. The constant acceleration is $\vec{a} = (3\hat{i} + 2\hat{j}) \text{ m/s}^2$.

The position vector of the particle at time $t$ is given by the kinematic equation:

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

Substituting the given values:

$$\vec{r}(t) = (0\hat{i} + 0\hat{j}) + (5\hat{i})t + \frac{1}{2}(3\hat{i} + 2\hat{j})t^2$$ $$\vec{r}(t) = 5t\hat{i} + \left(\frac{3}{2}t^2\hat{i} + \frac{2}{2}t^2\hat{j}\right)$$ $$\vec{r}(t) = \left(5t + \frac{3}{2}t^2\right)\hat{i} + t^2\hat{j}$$

The $x$-coordinate of the particle at time $t$ is $x(t) = 5t + \frac{3}{2}t^2$. We are given that the $x$-coordinate is $84 \text{ m}$ at some instant. Let this instant be $t$.

$$84 = 5t + \frac{3}{2}t^2$$

Multiply by 2 to clear the fraction:

$$168 = 10t + 3t^2$$

Rearrange into a standard quadratic form:

$$3t^2 + 10t - 168 = 0$$

We can solve this quadratic equation for $t$ using the quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:

$$t = \frac{-10 \pm \sqrt{10^2 - 4(3)(-168)}}{2(3)}$$ $$t = \frac{-10 \pm \sqrt{100 + 2016}}{6}$$ $$t = \frac{-10 \pm \sqrt{2116}}{6}$$

To find $\sqrt{2116}$: We can estimate it. $40^2=1600$, $50^2=2500$. Ends in 6, so it's $44^2$ or $46^2$. $46^2 = (40+6)^2 = 1600 + 480 + 36 = 2116$.

$$t = \frac{-10 \pm 46}{6}$$

We have two possible values for $t$: $t_1 = \frac{-10 + 46}{6} = \frac{36}{6} = 6 \text{ s}$ $t_2 = \frac{-10 - 46}{6} = \frac{-56}{6} = -\frac{28}{3} \text{ s}$ Since time cannot be negative, we take $t = 6 \text{ s}$.

Now, we need to find the speed of the particle at $t=6 \text{ s}$. The velocity vector of the particle at time $t$ is given by:

$$\vec{v}(t) = \vec{v}_0 + \vec{a}t$$

Substituting the given values and $t=6 \text{ s}$:

$$\vec{v}(t) = 5\hat{i} + (3\hat{i} + 2\hat{j})t$$ $$\vec{v}(6) = 5\hat{i} + (3\hat{i} + 2\hat{j})(6)$$ $$\vec{v}(6) = 5\hat{i} + 18\hat{i} + 12\hat{j}$$ $$\vec{v}(6) = (5+18)\hat{i} + 12\hat{j}$$ $$\vec{v}(6) = 23\hat{i} + 12\hat{j} \text{ m/s}$$

The speed of the particle at this time is the magnitude of the velocity vector:

$$\text{Speed} = |\vec{v}(6)| = \sqrt{(23)^2 + (12)^2}$$ $$\text{Speed} = \sqrt{529 + 144}$$ $$\text{Speed} = \sqrt{673} \text{ m/s}$$

We are given that the speed of the particle at this time is $\sqrt{\alpha} \text{ m/s}$. Comparing $\sqrt{673}$ with $\sqrt{\alpha}$, we find:

$$\alpha = 673$$

The final answer is $\boxed{673}$.