1
JEE Main 2024 (Online) 31st January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

The relation between time '$$t$$' and distance '$$x$$' is $$t=\alpha x^2+\beta x$$, where $$\alpha$$ and $$\beta$$ are constants. The relation between acceleration $$(a)$$ and velocity $$(v)$$ is :

A
$$a=-5 \alpha v^5$$
B
$$a=-3 \alpha v^2$$
C
$$a=-2 \alpha v^3$$
D
$$a=-4 \alpha v^4$$
2
JEE Main 2024 (Online) 29th January Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

A particle is moving in a straight line. The variation of position '$$x$$' as a function of time '$$t$$' is given as $$x=\left(t^3-6 t^2+20 t+15\right) m$$. The velocity of the body when its acceleration becomes zero is :

A
6 m/s
B
10 m/s
C
8 m/s
D
4 m/s
3
JEE Main 2024 (Online) 29th January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

A body starts moving from rest with constant acceleration covers displacement $$S_1$$ in first $$(p-1)$$ seconds and $$\mathrm{S}_2$$ in first $$p$$ seconds. The displacement $$\mathrm{S}_1+\mathrm{S}_2$$ will be made in time :

A
$$(2 p+1) s$$
B
$$(2 p-1) s$$
C
$$\left(2 p^2-2 p+1\right) s$$
D
$$\sqrt{\left(2 p^2-2 p+1\right)} s$$
4
JEE Main 2023 (Online) 15th April Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language
The position of a particle related to time is given by $x=\left(5 t^{2}-4 t+5\right) \mathrm{m}$. The magnitude of velocity of the particle at $t=2 s$ will be :
A
$14 \mathrm{~ms}^{-1}$
B
$16 \mathrm{~ms}^{-1}$
C
$10 \mathrm{~ms}^{-1}$
D
$06 \mathrm{~ms}^{-1}$
JEE Main Subjects
EXAM MAP