A particle starts from origin at $t = 0$ with a velocity $5\hat i\ \text{m s}^{-1}$ and moves in the $x$–$y$ plane under action of a force which 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 $84\ \text{m}$, then the speed of the particle at this time is $\sqrt\alpha\ \text{m s}^{-1}$. The value of $\alpha$ is ______.
Numerical value type. Enter your answer.
Answer: 673
Along $x$: $x = 5t + \frac12(3)t^2 = 84 \Rightarrow 3t^2 + 10t - 168 = 0$
$$t = \frac{-10 + \sqrt{100 + 2016}}{6} = \frac{-10+46}{6} = 6\ \text{s}$$
Velocity components: $v_x = 5 + 3\times6 = 23\ \text{m s}^{-1}$, $v_y = 2\times6 = 12\ \text{m s}^{-1}$
$$v^2 = 23^2 + 12^2 = 529 + 144 = 673 \;\Rightarrow\; \alpha = 673$$
Solution by Sreeraj P, M.Sc Physics