Question 1

A ball is thrown vertically downward with a velocity of  20 m/s  from the top of a tower. It hits the ground after some time with a velocity of  80 m/s . The height of the tower is :  (g = 10 m/s^2)

Accepted Answer

The initial velocity of the ball is  =20\ m/s .The final velocity of the ball is  =80\ m/s .The acceleration due to gravity is  =10\ m/s^2 .So, the height of the tower can be obtained from the expression: v^2=u^2+2gh Substitute the values in above expression: 80^2=20^2+(2	imes 10	imes h) h=\dfrac{6400-400}{20}\ m h=\dfrac{6000}{20}\Rightarrow 300\ m

Question 2

The speed of a swimmer in still water is 20 m/s. The speed of river water is 10 m/s and is flowing due east. If he is standing on the south bank and wishes to cross the river along the shortest path the angle at which he should make his strokes w.r.t north is given by :

Accepted Answer

V_{SR} = 20 m/s V_{RG} = 10 m/s \bar{V}_{SG} = \bar{V}_{SR} + \bar{V}_{RG} So, sin 	heta = \dfrac{|\bar{V}_{RG}|}{|\bar{V}_{SR}|} sin 	heta = \dfrac{10}{20}  \implies sin 	heta = \dfrac{1}{2} \implies 	heta = 30^o   west

Question 3

A particle moving with velocity  bar{V}  is acted by three forces shown by the vector triangle PQR. The velocity of the particle will :

Accepted Answer

As forces are forming closed loop in same order So,  \vec{F}_{net} = 0 \Rightarrow m \dfrac{d \vec{v}}{dt} = 0 \Rightarrow \vec{v}   = constant

Question 4

Preeti reached the metro station and found that the escalator was not working. She walked up the stationary escalator in time

t_{1}

. On other days, if she remains stationary on the moving escalator, then the escalator takes her up in time

t_{2}

. The time taken by her walk up on the moving escalator will be:

Accepted Answer

t_1=  time at stationary t_2=  time by escalator d  displacement of escalator v_1=  Velocity of stationary  v_2=  Velocity of Preeti (V_1+V_2) = \dfrac{d}{t'} , as  V_1=\dfrac{d}{t_1}, V_2 = \dfrac{d}{t_2} \dfrac{d}{t_1} +\dfrac{d}{t_2}=\dfrac{d}{t'} \rightarrow t' =\dfrac{t_1t_2}{t_1+t_2}

Question 5

The  x  and  y  coordinates of the particle at any time are  x=5t-2t^2  and  y=10t  respectively, where  x  and  y  are in meters and  t  in seconds. The acceleration of the particle at  t=2\ s  is:

Accepted Answer

x=5t-2t^2 y=10t \dfrac{dx}{dt}=5-4t a_x=\dfrac{d^2x}{dt^2}=-4 a_y=\dfrac{d^2y}{dt^2}=0 a=a_x+a_y=-4+0 a=-4 \ ms^{2}

Question 6

If the velocity of a particle is  v=At+Bt^2 , where  A  and  B  are constants, then the distance travelled by it between  1\ s  and  2\ s  is

Accepted Answer

We know that  v=\dfrac{dS}{dt} Integrating,  S=\int_{t=1}^{t=2}vdt=\int_1^2 (At+Bt^2)dt=A/2(2^2-1)+B/3 (2^3-1)=\dfrac{3}{2}A+\dfrac{7}{3}B

Motion In A Straight Line