QJEEM21Ph6a41: JEE Main 2021 Physics - Class 11 JEEM21Ph6a41 Solution: The velocity of a particle is v = v 0 + gt + Ft 2 . Its…

The velocity of a particle is given by $v = v_0 + gt + Ft^2$ . Its position is $x = 0$ at $t = 0$ . To find its displacement after time $t = 1$ , follow these steps:

Given:

$v = v_0 + gt + Ft^2$

We know that:

$\frac{dx}{dt} = v_0 + gt + Ft^2$

To find the displacement, integrate both sides with respect to $t$ :

$\int_{x = 0}^{x} dx = \int_{t = 0}^{t = 1} (v_0 + gt + Ft^2) \, dt$

This simplifies to:

$x = \left[ v_0 t + \frac{gt^2}{2} + \frac{Ft^3}{3} \right]_{t = 0}^{t = 1}$

Evaluating the integral from $t = 0$ to $t = 1$ :

$x = v_0 + \frac{g}{2} + \frac{F}{3}$

Therefore, the displacement after time $t = 1$ is:

$x = v_0 + \frac{g}{2} + \frac{F}{3}$