KCET 2026 | Motion in a Straight Line Question 2 | Physics

Two stones begin to fall from rest from the same height, with the second stone starting to fall ' $\Delta \mathrm{t}$ ' seconds after the first falls from rest. The distance of separation between the two stones becomes ' H ', ' $\mathrm{t}_0$ ' seconds after the first stone starts its motion. Then $\mathrm{t}_0$ is equal to

$\frac{\mathrm{H}}{\Delta \mathrm{t}}+\frac{\Delta \mathrm{t}}{2 \mathrm{~g}}$

$\frac{\mathrm{H}}{\mathrm{g} \Delta \mathrm{t}}-\frac{\Delta \mathrm{t}}{2}$

$\frac{\mathrm{H}}{\mathrm{g} \Delta \mathrm{t}}+\frac{\Delta \mathrm{t}}{2}$

$\frac{\mathrm{H}}{\mathrm{g} \Delta \mathrm{t}}$

KCET 2023

MCQ (Single Correct Answer)

-0

A body is moving along a straight line with initial velocity $$v_0$$. Its acceleration $$a$$ is constant. After $$t$$ seconds, its velocity becomes $$v$$. The average velocity of the body over the given time interval is

$$\bar{v}=\frac{v^2-v_0^2}{a t}$$

$$\bar{v}=\frac{v^2+v_0^2}{2 a t}$$

$$\bar{v}=\frac{v^2+v_0^2}{a t}$$

$$\bar{v}=\frac{v^2-v_0^2}{2 a t}$$

KCET 2022

MCQ (Single Correct Answer)

-0

The displacement $$x$$ (in $$\mathrm{m}$$) of a particle of mass $$m$$ (in $$\mathrm{kg}$$) moving in one dimension under the action of a force, is related to time $$t$$ (in sec) by $$t=\sqrt{x}+3$$. The displacement of the particle when its velocity is zero, will be

zero

6 m

2 m

4 m

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