JEE Physics 2024 Question with Solution

Given: Initial velocity in the $x$-direction is $u_x = 5 \, \text{m/s}$, acceleration components are $a_x = 3 \, \text{m/s}^2$ and $a_y = 2 \, \text{m/s}^2$, and the $x$-coordinate is $x = 84 \, \text{m}$.

Find: The value of $\alpha$ if speed $v = \sqrt{\alpha} \, \text{m/s}$.

Using motion in the $x$-direction,

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

So,

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

Now use

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

In the $y$-direction, initial velocity is zero, so

$$v_y = u_y + a_y t = 0 + 2 \cdot 6 = 12 \, \text{m/s}$$

Hence the speed is

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

Therefore, $\alpha = 673$.

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

Find: The numerical value of $\alpha$.

First determine the horizontal velocity from the kinematic relation:

$$v_x^2 - u_x^2 = 2a_x x$$ $$v_x^2 - 25 = 2(3)(84)$$ $$v_x^2 = 25 + 504 = 529$$ $$v_x = 23 \, \text{m/s}$$

Next find the time using horizontal motion:

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

Now calculate the vertical component of velocity:

$$v_y = 0 + a_y t = 2 \times 6 = 12 \, \text{m/s}$$

The resultant speed is

$$v = \sqrt{v_x^2 + v_y^2}$$ $$v = \sqrt{23^2 + 12^2} = \sqrt{673} \, \text{m/s}$$

So the required value is $673$.