Kinematics 1D- JEE Main Previous Year Questions with Solutions

JEE Main Previous Year Question of Physics with Solutions are available here. Practicing JEE Main Previous Year Papers Questions of Physics will help all the JEE aspirants in realizing the question pattern as well as help in analyzing their weak & strong areas.

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Previous Years AIEEE/JEE Mains Questions

Q. A particle has an initial velocity of $3 \hat{i}+4 \hat{j}$ and an acceleration of $0.4 \hat{\mathrm{i}}+0.3 \hat{\mathrm{j}}$. Its speed after 10s is :-

(1) 7 units

(2) 8.5 units

(3) 10 units

(4) $7 \sqrt{2}$ units

[AIEEE-2009]

Sol. (4)

v = u + at

u = 3i + 4j + (0.4 i + 0.3 j) × 10

= 3i + 4j + 4i + 3i

u = 7i + 7j

$|\overrightarrow{\mathrm{u}}|=\sqrt{7^{2}+7^{2}}$

$=7 \sqrt{2}$

Q. An object, moving with a speed of 6.25 m/s, is decelerated at a rate given by $\frac{\mathrm{dv}}{\mathrm{dt}}=-2.5 \sqrt{\mathrm{v}}$ where v is the instantaneous speed. The time taken by the object, to come to rest, would be :-

(1) 4 s (2) 8 s (3) 1 s (4) 2 s

[AIEEE-2011]

Sol. (4)

$\overrightarrow{\mathrm{a}}=\frac{\mathrm{d} \mathrm{v}}{\mathrm{dt}}$

$\frac{\mathrm{d} \mathrm{v}}{\mathrm{dt}}=-2.5 \sqrt{\mathrm{v}}$

$\int \mathrm{d} \mathrm{v} \times \mathrm{v}^{-1 / 2}=\int-2.5 \mathrm{dt}$

$\frac{\mathrm{v}^{\frac{1}{2}}}{\frac{1}{2}}=-2.5 \mathrm{t}$

$2 \sqrt{\mathrm{v}}=-2.5 \mathrm{t}+\mathrm{c}$

$2 \sqrt{6.25}=-2.5 \mathrm{t}+\mathrm{c}$

2 × 0.25 = c

5 = c

$2 \sqrt{\mathrm{v}}=-\mathrm{kt}+5$

0 = –kt + 5

kt = 5

$\mathrm{t}=\frac{5 \mathrm{sec}}{2.5}=2 \mathrm{sec}$

Q. From a tower of height H, a particle is thrown vertically upwards with a speed u. The time taken by the particle, to hit the ground, is n times that taken by it to reach the highest point of its path. The relation between H, u and n is :

(1) $2 \mathrm{g} \mathrm{H}=\mathrm{nu}^{2}(\mathrm{n}-2)$

(2) $\mathrm{g} \mathrm{H}=(\mathrm{n}-2) \mathrm{u}^{2}$

(3) $2 \mathrm{g} \mathrm{H}=\mathrm{n}^{2} \mathrm{u}^{2}$

(4) $\mathrm{g} \mathrm{H}=(\mathrm{n}-2)^{2} \mathrm{u}^{2}$

[JEE-MAIN-2014]

Sol. (1)

Time to reach highest point $=\mathrm{t}=\frac{\mathrm{u}}{\mathrm{g}}$

time to reach ground = nt

$\mathrm{S}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}$

$-\mathrm{H}=\mathrm{u}(\mathrm{nt})-\frac{1}{2} \mathrm{g}(\mathrm{nt})^{2}$ $\Rightarrow 2 \mathrm{gH}=\mathrm{nu}^{2}(\mathrm{n}-2)$

Q. A body is thrown vertically upwards. Which one of the following graphs correctly represent the velocity vs time ? [JEE Mains – 2017]

Sol. (1)

Velocity at any time t is

given by v = u + at

v = v0 + (–g)t

$\mathrm{v}=\mathrm{v}_{0}-\mathrm{gt}$

$\Rightarrow$ straight line with negative slope Q. All the graphs below are intended to represent the same motion. One of them does it incorrectly. Pick it up.  [JEE Mains – 2018]

Sol. (1)

In this question option (2) and (4) are the corresponding position – time graph and velocity –position graph of option (3) and its distance – time graph is given as Hence incorrect graph is option (1)

• July 10, 2020 at 8:58 am

fuckingly less questions

• July 8, 2020 at 6:18 pm

so less questions

• July 4, 2020 at 8:49 pm

Very less questions

• July 2, 2020 at 9:50 pm

Thank you for providing such a good questions
Please can you make a pdf of it.

• June 30, 2020 at 1:09 am

great effort

• June 22, 2020 at 3:14 pm

Best

• June 18, 2020 at 1:40 pm

SUPER

• June 16, 2020 at 10:30 am

too few questions

• June 14, 2020 at 12:43 pm

• May 27, 2020 at 10:10 am

Plz give more questions.

• May 27, 2020 at 8:23 am

Can’t u give more questions?

• May 24, 2020 at 4:48 pm

Good job

• May 20, 2020 at 8:40 pm

Plz give more questions..

• May 1, 2020 at 11:09 pm

• April 14, 2020 at 2:45 pm