MHT CET 2025 25th April Morning Shift | Motion Question 3 | Physics

Two boys are standing at points A and B on ground, where distance $\mathrm{AB}=\mathrm{x}$. The boy at B stars running perpendicular to $A B$ with velocity $\mathrm{v}_1$. The boy at A starts running simultaneously with velocity v and meets the other boy in time $t$. The value of $t$ is

$\left[\frac{x}{v-v_1}\right]^{1 / 2}$

$\left[\frac{x}{v_1-v}\right]^{1 / 2}$

$\left[\frac{x^2}{v^2-v_1^2}\right]^{1 / 2}$

$\left[\frac{x^2}{v_1^2-v^2}\right]^{1 / 2}$

MHT CET 2025 23rd April Evening Shift

MCQ (Single Correct Answer)

-0

A body when projected at an angle ' $\theta$ ' with the horizontal reaches a maximum height ' $H$ '. The time of flight of the body will be ( $\mathrm{g}=$ acceleration due to gravity)

$\frac{1}{2} \sqrt{\frac{2 \mathrm{H}}{\mathrm{g}}}$

$\sqrt{\frac{g}{2 H}}$

$2 \sqrt{\frac{2 \mathrm{H}}{\mathrm{g}}}$

$\sqrt{\frac{2 \mathrm{H}}{\mathrm{g}}}$

MHT CET 2025 23rd April Morning Shift

MCQ (Single Correct Answer)

-0

The acceleration (a) - time (T) graph for the body starting from rest is given below. The maximum speed of the body is

MHT CET 2025 23rd April Morning Shift Physics - Motion Question 5 English

$40 \mathrm{~m} / \mathrm{s}$

$80 \mathrm{~m} / \mathrm{s}$

$160 \mathrm{~m} / \mathrm{s}$

$200 \mathrm{~m} / \mathrm{s}$

MHT CET 2025 22nd April Evening Shift

MCQ (Single Correct Answer)

-0

At any time ' $t$ ', the co-ordinates of moving particle are $x=a t^2$ and $y=b t^2$. The speed of the particle is

$2 t \sqrt{a^2+b^2}$

$2 t \sqrt{a^2-b^2}$

$2 \mathrm{t}(\mathrm{a}+\mathrm{b})$

$\frac{2 t}{\sqrt{a^2+b^2}}$

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