A particle of mass m moves along a curve, y = x 2 . When the particle has x - coordinate as 1 2 m and x - component of velocity as 4 m s - 1 , then

the magnitude of velocity at that instant is, 4 2 m s - 1 .

the magnitude of velocity at that instant is, 2 2 m s - 1 .

The velocity of a particle moving in the positive direction of x - axis varies as v = 10 x . Assuming that at t = 0 , particle was at x = 0 ,

the acceleration of the particle is 50 m s - 2 .

the initial velocity of the particle is 2 . 5 m s - 1 .

the acceleration of the particle is 2 . 5 m s - 2 .

EASY

JEE Main

IMPORTANT

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If the velocity of a particle is given by $v = {(180 - 16 x)}^{1 / 2} {m s}^{- 1}$ , then its acceleration will be

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Important Questions on Motion in One Dimension

EASY

JEE Main

IMPORTANT

Chapterwise/Topicwise Daily Practice Problems (DPP) Mechanics I JEE Main & Advanced>Chapter 4 - Motion in One Dimension>DPP>Q 8

A particle moves along $x -$ axis as $x = 4 (t - 2) + a {(t - 2)}^{2}$ . Which of the following is true? Here $a$ is a positive constant.

EASY

JEE Main

IMPORTANT

Chapterwise/Topicwise Daily Practice Problems (DPP) Mechanics I JEE Main & Advanced>Chapter 4 - Motion in One Dimension>DPP>Q 9

The position vector of a particle is given as

\vec{r} = (t^{2} - 4 t + 6) \hat{i} + (t^{2}) \hat{j}

. The time after which the velocity vector and the acceleration vector become perpendicular to each other, is equal to

HARD

JEE Main

IMPORTANT

Chapterwise/Topicwise Daily Practice Problems (DPP) Mechanics I JEE Main & Advanced>Chapter 4 - Motion in One Dimension>DPP>Q 10

A particle of mass

m

moves along a curve,

y = x^{2}

. When the particle has

x -

coordinate as

\frac{1}{2} m

and

x -

component of velocity as

4 m s^{- 1}

, then

MEDIUM

JEE Main

IMPORTANT

Chapterwise/Topicwise Daily Practice Problems (DPP) Mechanics I JEE Main & Advanced>Chapter 4 - Motion in One Dimension>DPP>Q 11

The velocity of a particle moving in the positive direction of

x -

axis varies as

v = 10 \sqrt{x}

. Assuming that at

t = 0

, particle was at

x = 0

HARD

JEE Main

IMPORTANT

Chapterwise/Topicwise Daily Practice Problems (DPP) Mechanics I JEE Main & Advanced>Chapter 4 - Motion in One Dimension>DPP>Q 12

A particle moves along

x -

axis such that its position varies with time as

x = 50 t - 5 t^{2}

. Select the correct alternative(s).

MEDIUM

JEE Main

IMPORTANT

Chapterwise/Topicwise Daily Practice Problems (DPP) Mechanics I JEE Main & Advanced>Chapter 4 - Motion in One Dimension>DPP>Q 1

A particle is moving along a straight line with a constant acceleration of

- 4 m s^{- 2}

. It passes through a point

A

on the line with a velocity of

+ 8 m s^{- 1}

at some moment. The distance traveled by the particle in

5 \sec

after that moment will be

MEDIUM

JEE Main

IMPORTANT

Chapterwise/Topicwise Daily Practice Problems (DPP) Mechanics I JEE Main & Advanced>Chapter 4 - Motion in One Dimension>DPP>Q 2

A driver having a definite reaction time is capable of stopping his car over a distance of

30 m

on seeing a red traffic signal, when the speed of the car is

72 km h^{- 1}

and over a distance of

10 m

, when the speed is

36 km h^{- 1}

. Find the distance over which he can stop the car if it was running at a speed of

54 km h^{- 1}

. Assume that his reaction time and the deceleration of the car remains the same in all three cases.

EASY

JEE Main

IMPORTANT

Chapterwise/Topicwise Daily Practice Problems (DPP) Mechanics I JEE Main & Advanced>Chapter 4 - Motion in One Dimension>DPP>Q 3

A particle experiences a constant acceleration for

20 s

after starting from rest. If it travels a distance

S_{1}

in the first

10 s

and a distance

S_{2}

in the next

10 s

, then

If the velocity of a particle is given by v=(180-16x)1/2 m s-1, then its acceleration will be

Important Questions on Motion in One Dimension

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If the velocity of a particle is given by $v = {(180 - 16 x)}^{1 / 2} {m s}^{- 1}$ , then its acceleration will be