MHT CET 2025 26th April Evening Shift | Motion Question 1 | Physics

A car moving at a speed ' V ' is stopped in a certain distance when the breaks produce a deceleration ' $a$ '. If the speed of the car is ' $n v$ ', what must be the deceleration of the car to stop it in the same distance and in the same time?

$\sqrt{n} \cdot a$

$n \cdot a$

$n^2 \cdot a$

$\mathrm{n}^3 \cdot \mathrm{a}$

MHT CET 2025 26th April Morning Shift

MCQ (Single Correct Answer)

-0

A boy throws a ball vertically upwards from a bridge with velocity $5 \mathrm{~m} / \mathrm{s}$. It strikes water surface after 2 s . The height of the bridge is (Take $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$ )

20 m

15 m

12 m

10 m

MHT CET 2025 25th April Evening Shift

MCQ (Single Correct Answer)

-0

Two girls are standing at the ends ' $A$ ' and ' $B$ ' of a ground where $\mathrm{AB}=\mathrm{b}$. The girl at ' B ' starts running in a direction perpendicular to ' $A B$ ' with velocity ' $\mathrm{V}_1$ '. The girl at ' A ' starts running simultaneously with velocity ' $\mathrm{V}_2$ ' and in shortest distance meets the other girl in time ' $t$ '. The value of ' $t$ ' is

$\frac{b}{\sqrt{V_1{ }^2+V_2{ }^2}}$

$\frac{b}{V_1+V_2}$

$\frac{b}{V_2-V_1}$

$\frac{b}{\sqrt{V_2{ }^2-V_1{ }^2}}$

MHT CET 2025 25th April Morning Shift

MCQ (Single Correct Answer)

-0

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

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

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

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

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

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