The position of a particle is given by r → = i ^ 3 t - j ^ 2 t 2 + k ^ 4 , where t is in seconds and r → is in metres. Find the velocity and acceleration of the particle at the instant t = 2 s .

Position vector given in the question, r → = i ^ 3 t - j ^ 2 t 2 + k ^ 4 Now we know that we can get the velocity vector after differentiating the position vector i.e. v = d r → d t = d ( i ^ 3 t - j ^ 2 t 2 + k ^ 4 ) d t or, v = i ^ 3 - j ^ 4 t Putting value of t = 2 s we get, v = i ^ 3 - j ^ 8 So the magnitude of velocity will be, 3 2 + 8 2 = 73 and direction will be, θ = tan - 1 8 3 ≈ 70 ° with x -axis Similarly we can get, acceleration by differentiating the velocity vector. a = d v d t = d ( i ^ 3 - j ^ 4 t ) d t or, a = - j ^ 4 Hence, acceleration = - 4 m / s 2 along y -axis.

A particle starts from the origin at t = O S with a velocity j ^ 10 m / s and moves in the x - y plane with a constant acceleration i ^ 8 + j ^ 2 ms - 2 . At what time is the x component of the particle be 16 m ? What is the y coordinate of the particle at that time ?

It is given in the question that velocity of the particle is, v → = j ^ 10 m / s And the acceleration of the particle, a → = i ^ 8 + j ^ 2 ms - 2 Now we know that, acceleration can be obtained by differentiating the velocity. a → = d v → d t or, d v → = i ^ 8 + j ^ 2 d t Integrating the both side, v → ( t ) = i ^ 8 t + j ^ 2 t + u → Where, u → = Velocity Vector of the particle at t = 0 and v → = Velocity vector of the particle at time t But, v → = d r → d t d r → = v → d t = ( i ^ 8 t + j ^ 2 t + u → ) d t Integrating the equations with the conditions at t = 0 , r = 0 & t = t , r = r r → = u → t + 1 2 8 t 2 i ^ + 1 2 × 2 t 2 j ^ r → = u → t + 4 t 2 i ^ + t 2 j ^ r → = 10 j ^ t + 4 t 2 i ^ + t 2 j ^ or, x i ^ + y j ^ = 4 t 2 i ^ + 10 t + t 2 j ^ Since the motion of the particle is confined to the x - y plane, on equating the coefficients x = 4 t 2 y = 10 t + t 2 Now when x = 16 m t = 16 4 1 2 or, t = 2 s Now, y component, y = 10 × 2 + 2 2 = 24 m

HARD

11th West Bengal Board

IMPORTANT

Earn 100

A particle moving with uniform acceleration describes distances $s_{1}$ and $s_{2}$ in successive intervals of $t_{1}$ and $t_{2}$ seconds. Prove that the acceleration of the particle is $2 (s_{2} t_{1} - s_{1} t_{2}) / t_{1} t_{2} (t_{2} - t_{1})$ .

Important Questions on Motion of a Particle

HARD

11th West Bengal Board

IMPORTANT

A Text Book of PHYSICS PART I : CLASS 11>Chapter 5 - Motion of a Particle>EXERCISE>Q 18

A particle moves in a straight line with retardation

k ν^{2}

where

ν

is the velocity of the particle at time

t

and

k

is a positive constant. If

u

be the initial velocity and

x

be the displacement at time

t

show that

$\frac{1}{ν} = \frac{1}{u} + k t$

MEDIUM

11th West Bengal Board

IMPORTANT

A Text Book of PHYSICS PART I : CLASS 11>Chapter 5 - Motion of a Particle>EXERCISE>Q 18

A particle moves in a straight line with retardation

k ν^{2}

where

ν

is the velocity of the particle at time

t

and

k

is a positive constant. If

u

be the initial velocity and

x

be the displacement at time

t

show that

$x = \frac{1}{k} \log (1 + u k t)$ .

HARD

11th West Bengal Board

IMPORTANT

A Text Book of PHYSICS PART I : CLASS 11>Chapter 5 - Motion of a Particle>EXERCISE>Q 19

A stone is let fall from the top of a cliff. It describes last

h

height in time

t

second. Show that the total time of fall of the stone is

(\frac{t}{2} + \frac{h}{g t}) s

HARD

11th West Bengal Board

IMPORTANT

A Text Book of PHYSICS PART I : CLASS 11>Chapter 5 - Motion of a Particle>EXERCISE>Q 20

A ball of mass

100 g

is projected vertically upward from the ground with a velocity of

49 {ms}^{- 1}

. At the same time another identical ball is dropped from a height of

98 m

to fall freely along the same path as that followed by the first ball. After some time, the two balls collide and stick together and finally fall to the ground. Find the time of fall of the masses.

MEDIUM

11th West Bengal Board

IMPORTANT

A Text Book of PHYSICS PART I : CLASS 11>Chapter 5 - Motion of a Particle>EXERCISE>Q 21

On a two-lane road, car $A$ is travelling with a speed of

36 k m h^{- 1}

. Two cars $B$ and $C$ approach car

A

from opposite directions with a speed of

54 k m h^{- 1}

each. At a certain instant, when the distance

A B

is equal to

A C

, both being

1 km

B

decides to overtake $A$ before $C$ does. What minimum acceleration of car

B

is required to avoid an accident?

HARD

11th West Bengal Board

IMPORTANT

A Text Book of PHYSICS PART I : CLASS 11>Chapter 5 - Motion of a Particle>EXERCISE>Q 22

The position of a particle is given by

\vec{r} = \hat{i} 3 t - \hat{j} 2 t^{2} + \hat{k} 4

, where

t

is in seconds and

\vec{r}

is in metres.

Find the velocity and acceleration of the particle at the instant $t = 2 s$ .

HARD

11th West Bengal Board

IMPORTANT

A Text Book of PHYSICS PART I : CLASS 11>Chapter 5 - Motion of a Particle>EXERCISE>Q 23

A particle starts from the origin at

t = O S

with a velocity

\hat{j} 10 m / s

and moves in the

x - y

plane with a constant acceleration

(\hat{i} 8 + \hat{j} 2) {ms}^{- 2}

At what time is the $x$ component of the particle be $16 m$ ? What is the $y$ coordinate of the particle at that time ?

HARD

11th West Bengal Board

IMPORTANT

A Text Book of PHYSICS PART I : CLASS 11>Chapter 5 - Motion of a Particle>EXERCISE>Q 23

A particle starts from the origin at

t = 0 S

with a velocity

\hat{j} 10 m / s

and moves in the

x - y

plane with a constant acceleration

(\hat{i} 8 + \hat{j} 2) {ms}^{- 2}

What is the speed of the particle at that time $(t = 2 s)$ ?

A particle moving with uniform acceleration describes distances s1 and s2 in successive intervals of t1 and t2 seconds. Prove that the acceleration of the particle is 2s2t1-s1t2/t1t2t2-t1.

Important Questions on Motion of a Particle

Learn and Prepare for any exam you want !

A particle moving with uniform acceleration describes distances $s_{1}$ and $s_{2}$ in successive intervals of $t_{1}$ and $t_{2}$ seconds. Prove that the acceleration of the particle is $2 (s_{2} t_{1} - s_{1} t_{2}) / t_{1} t_{2} (t_{2} - t_{1})$ .