JEE Main 2024 — Motion In One Dimension Question with Solution

Given, $u_x=5 \mathrm{m} {\mathrm{s}}^{-1}, a_x=3 \mathrm{m} {\mathrm{s}}^{-2}, x=84 \mathrm{m}$

The formula to calculate the velocity of the particle along x-axis is given by

$v_x^2=u_x^2+2ax ...\left(1\right)$

From equation (1), it follows that

$$\begin{gathered} v_x^2=25+2\left(3\right)\left(84\right) \\ \Rightarrow v_{\mathrm{x}}=23 \mathrm{m} {\mathrm{s}}^{-1} \end{gathered}$$

Also, the velocity of the particle along x-axis can be written as

$v_x=u_x+a_{x}t ...\left(2\right)$

From equation (2), it follows that

$\begin{array}{rcl}t& =& \frac{23-5}{3}\\ & =& 6 \mathrm{s}\end{array}$

Similarly, for the $y-$ component of the velocity, it follows that

$\begin{array}{rcl}{v}_y& =& 0+a_{y}t\\ & =& 0+2\times \left(6\right)\\ & =& 12 \mathrm{m} {\mathrm{s}}^{-1}\end{array}$

Hence, the magnitude of the velocity is given by

$$\begin{gathered} v^2=v_x^2+v_y^2 \\ ={23}^2+{12}^2 \\ =673 \\ v=\sqrt{673} \mathrm{m} {\mathrm{s}}^{-1} \end{gathered}$$

Therefore, $\alpha =673$.