Question 144: Motion in 2D & Vectors

Question

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 _________.

Accepted Answer

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, $$\alpha = 673$$.

Motion in 2D & Vectors

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