KCET 2026 | Motion in a Straight Line Question 1 | Physics

The velocity of a particle moving along $x$-axis is given as $V = x^2 - 5x + 4$ (in m/s) where $x$ denotes the $x$-coordinate of the particle in metres. The magnitude of the acceleration of the particle when the velocity of the particle zero is

$2 \text{ m/s}^2$

$3 \text{ m/s}^2$

Zero

$1 \text{ m/s}^2$

KCET 2026

MCQ (Single Correct Answer)

-0

A car covers the first half of the distance between two places at $40$ km/h and another half at $50$ km/h. The average speed of the car is

$45.00$ km/h

$44.44$ km/h

$43.14$ km/h

$42.04$ km/h

KCET 2025

MCQ (Single Correct Answer)

-0

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

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

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

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

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

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