A particle starts from origin at $t=0$ with a velocity $5 \hat{i} \mathrm{~m} / \mathrm{s}$ and moves in $x-y$ plane under action of a force which produces a constant acceleration of $(3 \hat{i}+2 \hat{j}) \mathrm{m} / \mathrm{s}^2$. If the $x$-coordinate of the particle at that instant is $84 \mathrm{~m}$, then the speed of the particle at this time is $\sqrt{\alpha} \mathrm{~m} / \mathrm{s}$. The value of $\alpha$ is _________.

<p>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. <p>The motion of the particle in the $x$-direction can be described by the kinematic equation for uniformly accelerated motion:</p> <p>$x = x_0 + v_{0x}t + \frac{1}{2}a_xt^2$</p> <p>Given:</p> <p>$x_0 = 0 \mathrm{~m}$</p> <p>$v_{0x} = 5 \mathrm{~m/s}$</p> <p>$a_x = 3 \mathrm{~m/s}^2$</p> <p>$x = 84 \mathrm{~m}$ (at which we need to find the speed)</p> <p>Substituting these values into the kinematic equation:</p> <p>$$ 84 \mathrm{~m} = 0 \mathrm{~m} + (5 \mathrm{~m/s})t + \frac{1}{2}(3 \mathrm{~m/s}^2)t^2 $$</p> <p>Simplifying this equation, we get:</p> <p>$0 = \frac{3}{2}t^2 + 5t - 84$</p> <p>Using the quadratic formula to solve for $t$:</p> <p>$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$</p> <p>where $a = \frac{3}{2}, b = 5,$ and $c = -84$.</p> <p>$t = \frac{-5 \pm \sqrt{(5)^2 - 4(\frac{3}{2})(-84)}}{2(\frac{3}{2})}$</p> <p>$t = \frac{-5 \pm \sqrt{25 + 504}}{3}$</p> <p>$t = \frac{-5 \pm \sqrt{529}}{3}$</p> <p>$t = \frac{-5 \pm 23}{3}$</p> <p>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.</p> <p>$t = \frac{18}{3}$</p> <p>$t = 6 \mathrm{s}$</p> <p>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 $t = 6 \mathrm{s}$.</p> <p>The final velocity in the $x$-direction can be found using the formula for velocity with constant acceleration:</p> <p>$v_x = v_{0x} + a_xt$</p> <p>$v_x = 5 \mathrm{~m/s} + (3 \mathrm{~m/s}^2)(6 \mathrm{s})$</p> <p>$v_x = 5 \mathrm{~m/s} + 18 \mathrm{~m/s}$</p> <p>$v_x = 23 \mathrm{~m/s}$</p> <p>Similarly, for the $y$-direction:</p> <p>$v_y = v_{0y} + a_yt$</p> <p>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 $0$.</p> <p>$v_y = 0 + (2 \mathrm{~m/s}^2)(6 \mathrm{s})$</p> <p>$v_y = 12 \mathrm{~m/s}$</p> <p>Now we can compute the speed of the particle, which is the magnitude of the velocity vector:</p> <p>$v = \sqrt{v_x^2 + v_y^2}$</p> <p>$v = \sqrt{(23 \mathrm{~m/s})^2 + (12 \mathrm{~m/s})^2}$</p> <p>$v = \sqrt{529 + 144}$</p> <p>$v = \sqrt{673}$</p> <p>Therefore, the speed of the particle at the time its $x$-coordinate is $84 \mathrm{m}$ is $\sqrt{673} \mathrm{m/s}$.</p> <p>So, $\alpha = 673$.</p></p>