## The distance x covered by a particle in one dimensional motion varies with time t as x^2 = at^2 + 2 bt + c ….

Q: The distance x covered by a particle in one dimensional motion varies with time t as x2 = at2 + 2 bt + c . If the acceleration of particle depends on x as x-n , where n is an integer , the value of n is …….

Sol: x2 = at2 + 2 bt + c

$\large x = (at^2 + 2 bt + c)^{1/2}$

Differentiating with respect to time t ,

$\large \frac{dx}{dt} = \frac{1}{2}(at^2 + 2 bt + c)^{-1/2}(2at + 2b)$

Acceleration , $\displaystyle A = \frac{d^2x}{dt^2} = \frac{1}{2}[ (at^2 + 2 bt + c)^{-1/2}(2a) + (2at + 2b)(-\frac{1}{2})(at^2 + 2bt + c)^{-3/2}(2at + 2b)]$

$\large Ax = a – (\frac{at+b}{x})^2$

$\large Ax = \frac{ax^2 -(at+b)^2 }{x^2}$

$\large Ax = \frac{a(at^2 + 2bt + c)-(at+b)^2}{x^2}$

$\large A = \frac{ac-b^2}{x^3}$

$\large A \propto \frac{1}{x^3}$

$\large A \propto x^{-3}$

Hence , n = 3

## A balloon is moving up in air vertically above a point A on the ground . When it is at height h1…..

Q: A balloon is moving up in air vertically above a point A on the ground . When it is at height h1 , a girl standing at a distance d as shown in figure . sees it at an angle 45° with respect to vertical . When the balloon climbs up a further height h2 , it is seen at an angle 60° with respect to vertical if the girl moves further by a distance 2.464 d . Then height h2 is (given tan30° = 0.5774)

(a) 1.464 d

(b) 0.732 d

(c) 0.464 d

(d) d

Ans: (d)

Sol:From figure , $\large \frac{h_1}{d} = tan45^o$

$\large \frac{h_1}{d} = 1$

h1 = d

$\large \frac{h_1+h_2}{d+2.464 d} = tan30^o$

$\large \frac{h_1+h_2}{d+2.464 d} = 0.5744$

h1 + h2 = 2 d

h2 = d

## A tennis ball is released from height d and after freely falling on a wooden floor it rebounds and reaches height d/2 ….

Q: A tennis ball is released from height d and after freely falling on a wooden floor it rebounds and reaches height d/2 . The velocity versus height of the ball during its motion may be represented graphically by

Click to See Answer :
Ans: (a)

Sol: When ball is falling down from height d and reaches at height h , then velocity acquired by the ball is

v2= 2g(d-h)    ….(i)

This is downward velocity and is taken as negative .

When h = d , v = 0

When h =0 , v2 = (2 g d)

Hence from eqn(i) , the variation of v & h is parabolic below h axis

## Train A and Train B are running on parallel track in the opposite direction with speed of 36 km/h and 72 km/h respectively ….

Q: Train A and Train B are running on parallel track in the opposite direction with speed of 36 km/h and 72 km/h respectively . A person is walking in train A in the direction opposite to its motion with a speed of 1.8 km/h . Speed ( in m/s ) of this person as observed from train B will be close to ( take the distance between the tracks are negligible )

(a) 29.5 m/s

(b) 28.5 m/s

(c) 31.5 m/s

(d) 30.5 m/s

Ans: (a)

Sol: Let train A is moving along x-axis .

Velocity of train A with respect to ground , $\displaystyle \vec{v_A} = 36 \hat{i} \; km/h$

Velocity of train B with respect to ground , $\displaystyle \vec{v_B} = 72 (-\hat{i}) \; km/h$

Velocity of person with respect to train A , $\displaystyle \vec{v_{PA}} = -1.8 \hat{i} \; km/h$

Velocity of person with respect to train B , $\displaystyle \vec{v_{PB}} = \vec{v_P} – \vec{v_B}$

$\displaystyle \vec{v_{PB}} = \vec{v_{PA}} + \vec{v_A} – \vec{v_B}$

$\displaystyle \vec{v_{PB}} = -1.8 \hat{i} + 36 \hat{i} – (-72 \hat{i})$

$\displaystyle \vec{v_{PB}} = 106.2 \hat{i} \; km/h$

$\displaystyle \vec{v_{PB}} = 106.2 \hat{i} \times \frac{5}{18} \; m/s$

= 29.5 m/s

## A particle starts from origin at t= 0 with an initial velocity of 3.0 i m/s and moves in x-y plane with a constant acceleration ….

Q: A particle starts from origin at t= 0 with an initial velocity of $3.0 \hat{i} \; m/s$ and moves in x-y plane with a constant acceleration $\displaystyle (6 \hat{i} + 4 \hat{j}) m/s^2$  . The x-co-ordinate of the particle at the instant when its y-co-ordinate is 32 m is D meter . The value of D is

(a) 50

(b) 40

(c) 32

(d) 60

Ans: (d)

Sol: $S_y = u_y t + \frac{1}{2}a_y t^2$

$32 = 0 + \frac{1}{2}(4) t^2$

$32 = 2 t^2$

t = 4 sec

$S_x = u_x t + \frac{1}{2}a_x t^2$

$D = 3 \times 4 + \frac{1}{2}(6) 4^2$ (since u_x = 3 m/s)

D = 12 + 48 = 60 m