A balloon starts rising from the ground with an acceleration of $1.25\,\,m{\rm{/}}{s^2}$ after 8 seconds, a stone released from the balloon, the stone will $[g = 10\,\,m{\rm{/}}{s^2}]$

begin to move down after being released

If position vector of a particle at time $t$ is $\vec{r}=3 t \hat{i}+(8-4 t) \hat{j} \mathrm{~m}$, then

Particle is moving with variable velocity in straight line

Particle moving is in parabolic path

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JEE • PHYSICS • MOTION IN A STRAIGHT LINE

Motion in a straight line practice · Hard

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Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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1 Hard

1 of 16

Analyzing +4 | -1

A target is made of two plates, one of wood and the other of iron. The thickness of the wooden plate is 4 cm and that of iron plate is 2 cm. A bullet fired goes through the wood first and then penetrates 1 cm into iron. A similar bullet fired with the same velocity from opposite direction goes through iron first and then penetrates 2cm into wood. If

{a_1}

and

{a_2}

be the retardation offered to the bullet by wood and iron plates respectively, then

View answer and explanation

Correct option: B

{a_2} = 2{a_1}

Option B (

{a_2} = 2{a_1}

) is correct because the retardation offered by iron is twice that of wood, derived from the equations of motion.

2 Hard

2 of 16

Application +4 | -1

The relation

3t = \sqrt 3 x + 6

describes the displacement of a particle in one direction, where x is in meter and t in second. The displacement when velocity is zero

View answer and explanation

Correct option: D

zero

Option D (zero) is correct because when the velocity is zero, the displacement is also zero based on the given equation.

3 Hard

3 of 16

Analyzing +4 | -1

A point moves with a uniform acceleration and

v_1, v_2, v_3

denote the average velocities in three successive intervals of time

t_1, t_2, t_3

. Which of the following relations is correct?

View answer and explanation

Correct option: B

\left(v_1-v_2\right):\left(v_2-v_3\right)=\left(t_1+t_2\right):\left(t_2+t_3\right)

The correct answer is B because the relationship between average velocities and time intervals with uniform acceleration leads to the given ratio.

4 Hard

4 of 16

Application +4 | -1

A particle moves according to the equation

\frac{d v}{d t}=\alpha-\beta v

, where

\alpha

and

\beta

are constants. Find velocity as a function of time. Assume, body starts from rest

View answer and explanation

Correct option: B

v=(\alpha / \beta)\left(1-e^{-\beta t}\right)

Option B (

v=(\alpha / \beta)(1-e^{-\beta t})

) is correct because it is the solution to the given differential equation with the initial condition that the particle starts from rest.

5 Hard

5 of 16

Application +4 | -1

A body

P

is thrown vertically up with velocity 30 m/s and another body

Q

is thrown up along the same vertically line with the same velocity but 1 second later from the ground. When they meet

\left( {g = 10\,m/{s^2}} \right)

View answer and explanation

Correct option: C

P travels for 3.5 s

Option C (P travels for 3.5 s) is correct because it's calculated by finding the time when the two projectiles meet, considering the 1-second delay.

6 Hard

6 of 16

Application +4 | -1

A balloon starts rising from the ground with an acceleration of

1.25\,\,m\mathrm{/}{s^2}

after 8 seconds, a stone released from the balloon, the stone will

[g = 10\,\,m\mathrm{/}{s^2}]

View answer and explanation

Correct option: A

reach the ground 4 s

Option A (reach the ground 4 s) is correct because the time taken for the stone to reach the ground is calculated using kinematic equations of motion.

7 Hard

7 of 16

Application +4 | -1

The co-ordinates of a moving particle any time “

t

” are given by

x = \alpha {t^3}\,\,\& \,\,y = \beta {t^3}.

The speed of the particle at time

t = 2\,\,\sec

View answer and explanation

Correct option: C

12\sqrt {{\alpha ^2} + {\beta ^2}}

Option C (

12\sqrt {{\alpha ^2} + {\beta ^2}}

) is correct because the speed is found by differentiating the position functions with respect to time and then finding the magnitude of the velocity vector at t=2.

8 Hard

8 of 16

Application +4 | -1

A ball dropped from a point

\mathrm{A}

falls down vertically to

\mathrm{C}

, through the midpoint

\mathrm{B}

. The descending time from

\mathrm{A}

\mathrm{B}

and that from

\mathrm{A}

\mathrm{C}

are in the ratio

View answer and explanation

Correct option: D

1: \sqrt{2}

Option D (

1: \sqrt{2}

) is correct because the time taken is proportional to the square root of the distance.

9 Hard

9 of 16

Application +4 | -1

Amongst the following equation of motion, which represents uniformly accelerated motion?

View answer and explanation

Correct option: C

t=\sqrt{\frac{x+a}{b}}

Option C (

t=\sqrt{\frac{x+a}{b}}

) is correct because it can be rearranged to show that displacement x is proportional to the square of time t, which is the characteristic of uniformly accelerated motion.

10 Hard

10 of 16

Application +4 | -1

The deceleration experienced by a moving motorboat after its engine is cut off, is given by

\mathrm{dv} / \mathrm{dt}=-\mathrm{kv}^3

where

\mathrm{k}

is constant. If

\mathrm{v}_0

is the magnitude of the velocity at cut-off, the magnitude of the velocity at a time

t

after the cut-off is

View answer and explanation

Correct option: A

\frac{\mathrm{v}_0}{\sqrt{\left(2 \mathrm{v}_0{ }^2 \mathrm{kt}+1\right)}}

Option A (

\frac{{\mathrm{v}_0}}{{\sqrt{{\left(2 \mathrm{v}_0{ }^2 \mathrm{kt}+1\right)}}}}

) is correct because it is derived by solving the given differential equation with the initial condition.

11 Hard

11 of 16

Application +4 | -1

The velocity vector of a particle moving over a horizontal plane is given by

\vec{v}=2 t \hat{i}

. The ratio of the magnitude of instantaneous acceleration of the particle at

{t}=2 \mathrm{sec}

to the magnitude of displacement of the particle during the interval of first

2 \mathrm{sec}

View answer and explanation

Correct option: A

3/4

Option A (3/4) is correct because it represents the ratio of the magnitude of instantaneous acceleration to the magnitude of displacement, calculated using the given velocity vector.

12 Hard

12 of 16

Analyzing +4 | -1

The distance x covered by a particle in one-dimensional motion varies with time t as

{x^2} = a{t^2} + 2bt + c

. The accelerating of the particle varies as

View answer and explanation

Correct option: A

{x^{ - 3}}

Option A (

{x^{ - 3}}

) is correct. By differentiating the given equation twice with respect to time, the acceleration is found to be inversely proportional to x cubed.

13 Hard

13 of 16

Application +4 | -1

The ratio of distances traversed in successive intervals of time when a body falls freely under gravity from certain height is

View answer and explanation

Correct option: C

1: 3: 5

Option C (

1: 3: 5

) is correct because the distances traversed in successive equal intervals of time are in the ratio of odd integers.

14 Hard

14 of 16

Analyzing +4 | -1

A stone dropped from a building of height h and it reaches after t seconds on earth. From the same building if two stones are thrown (one upwards and other downwards) with the same velocity u and they reach the earth surface after

{t_1}

and

{t_2}

seconds respectively, then

View answer and explanation

Correct option: C

t = \sqrt {{t_1}{t_2}}

Option C (

t = \sqrt {{t_1}{t_2}}

) is correct because it represents the geometric mean of the times when the stones are thrown upwards and downwards, derived from kinematic equations.

15 Hard

15 of 16

Analyzing +4 | -1

If position vector of a particle at time

t

\vec{r}=3 t \hat{i}+(8-4 t) \hat{j} \mathrm{~m}

, then

View answer and explanation

Correct option: A

Particle is moving with constant velocity in straight line

The correct answer is A because the velocity vector is constant, indicating motion in a straight line with constant velocity.

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16 Hard

16 of 16

Application +4 | -1

A ball dropped from a point

P

crosses a point

Q

t

seconds. The time taken by it to travel from Q to R, if

PQ = QR

View answer and explanation

Correct option: D

\left( {\sqrt 2 - 1} \right)t

Option D (

\left( {\sqrt 2 - 1} \right)t

) is correct because it is the time taken to travel from Q to R, calculated using kinematic equations for constant acceleration.

Revision notes and key takeaways

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Key Takeaways

✓ This set contains 16 single-correct-answer MCQs on MOTION IN A STRAIGHT LINE.
✓ Try each question before opening its answer. Read the explanation where one is available.
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