KCET 2025 | Motion in a Straight Line Question 1 | 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

KCET 2021

MCQ (Single Correct Answer)

-0

For a body moving along a straight line, the following $$v$$-$$t$$ graph is obtained.

KCET 2021 Physics - Motion in a Straight Line Question 7 English

According to the graph, the displacement during

uniform acceleration is greater than that during uniform motion

uniform acceleration is less than that during uniform motion

uniform acceleration is equal to that during uniform motion

uniform motion is zero

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