QJEEM24Ph3acb: JEE Main 2024 Physics - Class 11 JEEM24Ph3acb Solution: A particle starts from origin at with a velocity and moves in…

To solve for the value of $\alpha$ , which represents the square of the speed of the particle at the time its $x$ -coordinate is $84 \mathrm{m}$ , we need to first determine the time at which the particle reaches this $x$ -coordinate, and then use this time to calculate its final velocity in both the $x$ and $y$ directions.

The motion of the particle in the $x$ -direction can be described by the kinematic equation for uniformly accelerated motion:

$x = x_0 + v_{0x}t + \frac{1}{2}a_xt^2$

Given:

$x_0 = 0 \mathrm{~m}$

$v_{0x} = 5 \mathrm{~m/s}$

$a_x = 3 \mathrm{~m/s}^2$

$x = 84 \mathrm{~m}$ (at which we need to find the speed)

Substituting these values into the kinematic equation:

$84 \mathrm{~m} = 0 \mathrm{~m} + (5 \mathrm{~m/s})t + \frac{1}{2}(3 \mathrm{~m/s}^2)t^2$

Simplifying this equation, we get:

$0 = \frac{3}{2}t^2 + 5t - 84$

Using the quadratic formula to solve for $t$ :

$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

where $a = \frac{3}{2}, b = 5,$ and $c = -84$ .

$t = \frac{-5 \pm \sqrt{(5)^2 - 4(\frac{3}{2})(-84)}}{2(\frac{3}{2})}$

$t = \frac{-5 \pm \sqrt{25 + 504}}{3}$

$t = \frac{-5 \pm \sqrt{529}}{3}$

$t = \frac{-5 \pm 23}{3}$

In this scenario, since we're looking for a time when the particle reaches $84 \mathrm{m}$ , we only consider the positive root because time cannot be negative.

$t = \frac{18}{3}$

$t = 6 \mathrm{s}$

Now we have the time at which the particle's $x$ -coordinate is $84 \mathrm{m}$ . Next, we find final velocities in $x$ and $y$ directions at $t = 6 \mathrm{s}$ .

The final velocity in the $x$ -direction can be found using the formula for velocity with constant acceleration:

$v_x = v_{0x} + a_xt$

$v_x = 5 \mathrm{~m/s} + (3 \mathrm{~m/s}^2)(6 \mathrm{s})$

$v_x = 5 \mathrm{~m/s} + 18 \mathrm{~m/s}$

$v_x = 23 \mathrm{~m/s}$

Similarly, for the $y$ -direction:

$v_y = v_{0y} + a_yt$

Since the particle starts from the origin and is only subject to a force after $t=0$ , its initial velocity in the $y$ -direction is $0$ .

$v_y = 0 + (2 \mathrm{~m/s}^2)(6 \mathrm{s})$

$v_y = 12 \mathrm{~m/s}$

Now we can compute the speed of the particle, which is the magnitude of the velocity vector:

$v = \sqrt{v_x^2 + v_y^2}$

$v = \sqrt{(23 \mathrm{~m/s})^2 + (12 \mathrm{~m/s})^2}$

$v = \sqrt{529 + 144}$

$v = \sqrt{673}$

Therefore, the speed of the particle at the time its $x$ -coordinate is $84 \mathrm{m}$ is $\sqrt{673} \mathrm{m/s}$ .

So, $\alpha = 673$ .