Question

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

Answer 1

Given initial velocity $u_x = 5 \, \text{m/s}$, acceleration $a_x = 3 \, \text{m/s}^2$, and $x = 84 \, \text{m}$.

Using the equation:

$v_x^2 = u_x^2 + 2a_xx$

$v_x^2 = 5^2 + 2 \cdot 3 \cdot 84 = 25 + 504 = 529$

$v_x = 23 \, \text{m/s}$

Similarly, for the y-direction:

$v_y = u_y + a_y t = 0 + 2 \cdot t$

Using $v_x = u_x + a_x t$:

$t = \frac{v_x - u_x}{a_x} = \frac{23 - 5}{3} = 6 \, \text{s}$

Then,

$v_y = 2 \times 6 = 12 \, \text{m/s}$

The speed of the particle is:

$v = \sqrt{v_x^2 + v_y^2} = \sqrt{23^2 + 12^2} = \sqrt{529 + 144} = \sqrt{673}$

Thus, $\alpha = 673$.