QJEEM20Ph29a0: JEE Main 2020 Physics - Class 11 JEEM20Ph29a0 Solution: A particle is moving along the x-axis with its coordinate with…

For a particle ‘A’, its position along the x-axis as a function of time $t$ is given by:

$x(t) = 10 + 8t - 3t^2$

To find the velocity $v_A$ , we take the derivative of $x(t)$ with respect to $t$ :

$v_A = \frac{d}{dt}[10 + 8t - 3t^2] = 8 - 6t$

At $t = 1$ second, the velocity of particle A is:

$\vec{v_A} = (8 - 6 \cdot 1)\hat{i} = 2\hat{i}$

For a particle ‘B’, its position along the y-axis as a function of time $t$ is given by:

$y(t) = 5 - 8t^3$

To find the velocity $v_B$ , we take the derivative of $y(t)$ with respect to $t$ :

$v_B = \frac{d}{dt}[5 - 8t^3] = -24t^2$

At $t = 1$ second, the velocity of particle B is:

$\vec{v_B} = -24 \cdot 1^2 \hat{j} = -24\hat{j}$

The velocity of particle B relative to particle A ( $\vec{v_{B/A}}$ ) is calculated as:

$\vec{v_{B/A}} = \vec{v_B} - \vec{v_A}$

$\vec{v_{B/A}} = -24\hat{j} - 2\hat{i}$

To find the magnitude of $\vec{v_{B/A}}$ :

$|\vec{v_{B/A}}| = \sqrt{(-24)^2 + (-2)^2} = \sqrt{576 + 4} = \sqrt{580}$

Therefore, $v$ is:

$v = 580$