Q: A ball is thrown vertically upwards. Which of the following plots represents the speed-time graph of the ball during its flight if the air resistance is not ignored ?

\textbf{Step} : For a body which is going upwards then v=ugt Slope of the graph \dfrac{dv}{dt}=-g (constant) During upward motion, the gravity on object and air resistance will oppose its motion because of which the speed of the body decreases and becomes zero at highest point. Later the body moves downwards because of the gravity pull but air resistance opposes and the velocity of body increases.But when we take into account the effect of resistance it will have sharper slope.Thus option D is correct.

Question 1

The co-ordinates of a moving particle at any time

t

are given by

x = α t^{3}

and

y = β t^{3}

. The speed to the particle at time

t

is given by

Accepted Answer

x = \alpha t^3 \,\,\, ; \,\,\, y = \beta t^3 \dfrac{dx}{dt} = \alpha (3t^2) \,\, ; \,\, \dfrac{dy}{dt} = \beta (3t^2) v_x = 3 \alpha t^2 \,\, ; \,\, v_y = 3 \beta t^2 resultant velocity  = \sqrt{v_x^2 + v_y^2}                               = 3t^2 \sqrt{\alpha^2 + \beta^2}

Question 2

Two particles A and B move with velocities  v_1\,and\, v_2   respectively along the x and y axis.  The initial separation between them is 'd' as shown in the figure. Find the least distance between them during their motion

Accepted Answer

Let  \vec{v_{1}} = v_{1} \hat i \vec{v_{2}} = v_{2} \hat j \vec{v_{2}} - \vec{v_{1}} = v_{2} \hat i - v_{1} \hat j \vec{s_2}(t)-\vec{s_1}(t) = d \hat j + v_2 t \hat i - v_1 t \hat j |\vec{s_2}(t) - \vec{s_1}(t)|^2 = (v_2 t)^2 +(d-v_1t)^2                               =(v_2^2+v_1^2)t^2 - 2dv_1t + d^2                               = \displaystyle{(\sqrt{v_1^2+v_2^2}  t - \dfrac{dv_1}{\sqrt{v_2^2+v_1^2}})^2 + d^2 - \dfrac{d^2v_1^2}{v_1^2+v_2^2} }                               \geq \dfrac{d^2 v_2^2}{v_1^2+v_2^2} |\vec {s_1} (t) - \vec{s_2}(t)| \geq \dfrac{dv_2}{\sqrt{v_1^2+v_2^2}}

Question 3

Which of the following position-time graphs shows an object that speeds up at first, travels at a constant velocity for a while, and then slows down?

Accepted Answer

Slope of position time graph gives velocity.So initial increasing and slope then constant the decreasing slope will be the required anser. In option A,B,E slopes are always constant these these cannot be possible.In option C initial slope is decreasing then constant then increasing.But in option D initially slpoe is increasing and then constant and then decreasing so this option is best for the statement required.

Question 4

A swimmer crosses a river with minimum possible time

10

second. And when he reaches the other end starts swimming in the direction towards the point from where he started swimming. Keeping the direction fixed the swimmer crosses the river in

15

sec. The ratio of speed of swimmer with respect to water and the speed of river flow is:

(Assume constant speed of river & swimmer)

Accepted Answer

A swimmer crosses  v   =  velocity of man w.r.t . river u   =  velocity of river A\overset{t}{ightarrow}B=\dfrac{d}{v}\Rightarrow 10=\dfrac{d}{v}\Rightarrow d=10 Vightarrow (1) B\overset{t}{ightarrow}C=\dfrac{d}{v cos 	heta}\Rightarrow 15=\dfrac{d}{v cos 	heta} \Rightarrow d =15 v cos 	heta ightarrow (2) (1) & (2)  \Rightarrow cos 	heta =\dfrac{2}{3} \Rightarrow sec 	heta = \dfrac{3}{2}. \because tan 	heta =\dfrac{u}{v}\ 	herefore \sqrt{sec^{2} 	heta -1}=\dfrac{u}{v} \displaystyle\Rightarrow \frac{u}{v}=\sqrt{\frac{9}{4} -1}=\frac{\sqrt 5}{2}\ \Rightarrow \dfrac{v}{u}=\dfrac{2}{\sqrt{5}}

Question 5

A particle moves with constant acceleration along a straight line starting from rest. The percentage increase in its displacement during the 4th second compared to that in the 3rd second is:

Accepted Answer

The displacement of the particle in  nth   second           S_n = u+  \dfrac{1}{2}a (2n-1) where  all the symbols have their usual meaning.Given :       u = 0 \implies  S_n  = \dfrac{1}{2} a(2n-1) Thus displacement in the  3rd  second              S_{3}  = \dfrac{1}{2}a (2	imes 3 - 1)   =\dfrac{5a}{2} Displacement in the  4th  second              S_{4}  = \dfrac{1}{2}a (2	imes 4 - 1)   =\dfrac{7a}{2} Percentage increase in the displacement      \dfrac{\Delta S}{S_3} 	imes 100  = \dfrac{S_4 - S_3}{S_3} 	imes 100    \dfrac{\Delta S}{S_3} 	imes 100  = \dfrac{\frac{7a}{2}  -\frac{5a}{2}}{\frac{5a}{2}} 	imes 100  = 40 Thus percentage increase in the displacement is  40  %

Question 6

A ball is dropped vertically from height  h  and is bouncing elastically on the floor (see figure). Which of the following plots best depicts the acceleration of the ball as a function of time.

Accepted Answer

During free fall of the ball, gravitational force acts on the body in downward direction always and so the acceleration of the ball remains in vertically downward direction which is constant with time. Its acceleration becomes positive only at single instant of time during the collision of ball with the ground. The ground imparts an upward impulse and hence the upward acceleration on the ball which makes the ball to move upward, but as soon as the ball looses contact with the ground, its accelerates in the downward direction again. Hence, option B is correct.

Question 7

A ball is thrown vertically upwards. Which of the following plots represents the speed-time graph of the ball during its flight if the air resistance is not ignored ?

Accepted Answer

extbf{Step} : For a body which is going upwards then  v=ugt Slope of the graph  \dfrac{dv}{dt}=-g (constant) During upward motion, the gravity on object and air resistance will oppose its motion because of which the speed of the body decreases and becomes zero at highest point. Later the body moves downwards because of the gravity pull but air resistance opposes and the velocity of body increases.But when we take into account the effect of resistance it will have sharper slope.Thus option D is correct.

Question 8

Two particles, 1 and 2, move with constant velocities

v_{1}

and

v_{2}

along two mutually perpendicular straight lines toward the intersection point

O

. At the moment

t = 0

the particles were located at the distances

l_{1}

and

l_{2}

from the point

O

. At time

t_{m} = x \frac{v _{1} l _{1} + v _{2} l _{2}}{v _{1}^{2} + v _{2}^{2}}

, the distance between the particles become the smallest. Value of

x

will be

Accepted Answer

Let the particle 1 and 2 be at points  B  and  A  at  t=0  at the distances  l_1  and  l_2  from intersection point  O .Let us fix the inertial frame with the particle 2. Now the particle 1 moves in relative to this reference frame with a relative velocity  \vec{v_{12}}=\vec{v_1}-\vec{v_2} , and its trajectory is the straight line  BP . Obviously, the minimum distance between the particles is equal to the length of the perpendicular  AP  dropped from point  A  on to the straight line  BP  (figure shown below).From the figure shown below,  v_{12}=\sqrt{v_1^2+v_2^2} , and  \displaystyle	an{	heta}=\frac{v_1}{v_2}  (1)The shortest distance AP=AM\sin{	heta}=(OA-OM)\sin{	heta}=(l_2-l_1\cot{	heta})\sin{	heta} or  \displaystyle AP=\left(l_2-l_1\frac{v_2}{v_1}ight)\frac{v_1}{\sqrt{v_1^2+v_2^2}}=\frac{v_1l_2-v_2l_1}{\sqrt{v_1^2+v_2^2}}  (using 1)The sought time can be obtained directly from the condition that  {(l_1-v_1t)}^2+{(l_2-v_2t)}^2  is minimum. This gives  \displaystyle t=\frac{l_1v_1+l_2v_2}{v_1^2+v_2^2} .

Question 9

A motorboat going downstream overcame a raft at a point  A ;  	au = 60\:min  later it turned back and after some time the raft is at a distance  l = 6.0\:km  from the point  A . Find the flow velocity in  km/h  assuming the duty of the engine to be constant.

Accepted Answer

Considering the given figure below.Let  v_0  be the stream velocity and  v^\prime  the velocity of motorboat with respect to water. The motorboat reached point  B  while going downstream with velocity  (v_0+v^\prime)  and then returned with velocity  (v^\prime-v_0)  and passed the raft at point  C . Let  t  be the time for the raft (which flows with stream with velocity  v_0 ) to move from point  A  to  C , during which the motorboat moves from  A  to  B  and then from  B  to  C .Therefore \displaystyle\frac{l}{v_0}=	au+\frac{(v_0+v^\prime)	au-l}{(v^\prime-v_0)} On solving, we get  \displaystyle v_0=\frac{l}{2	au}=3  km/hr .

Question 10

Three points are located at the vertices of an equilateral triangle whose side equals  a . They all start moving simultaneously with velocity  v  constant in modulus, with the first point heading continually for the second, the second for the third, and the third for the first. The time at which the points converge is given as  \displaystyle t=\frac{2a}{xv} . Find  x .

Accepted Answer

From the symmetry of the problem all the three points are always located at the vertices of equilateral triangles of varying side length and finally meet at the centroid of the initial equilateral triangle whose side length is  a , in the sought time interval (say  t ).Let us consider an arbitrary equilateral triangle of edge length  l  (say).Then the rate by which 1 approaches 2, 2 approaches 3, and 3 approaches 1, becomes: \displaystyle\frac{-dl}{dt}=v-v\cos{\left(\frac{2\pi}{3}\right)} On integrating:  \displaystyle-\int_a^0{dl}=\frac{3v}{2}\int_0^t{dt} \displaystyle a=\frac{3}{2}vt  so  \displaystyle t=\frac{2a}{3v}

Question 11

A man wants to reach from A to the opposite corner of the square C (Fig) . The sides of the square are 100 m. A central square of 50m is filled with sand. Outside this square, he can walk at a speed 1 m/s. In the central square, he can walk only at a speed of v m/s (v < 1). What is smallest value of v for which he can reach faster via a straight path through the sand than any path in the square outside the sand?

Accepted Answer

AC = \sqrt{(100)^{2} + (100)^{2}}  =  100\sqrt{2}m PQ = \sqrt{(50)^{2} + (50)^{2}}  =  50\sqrt{2} m AP = \dfrac{AC-PQ}{2} = \dfrac{100\sqrt{2}- 50\sqrt{2}}{2}  =  25\sqrt{2}m RC = AR= 25\sqrt{2}m Consider the straight-line path  APQC  through the sand. Time is taken to go from  A  to  C  via this path T_{sand}  = \dfrac{AP + QC}{1}+ \dfrac{PQ}{v} =  \dfrac{25\sqrt{2} + 25\sqrt{2}}{1} +\dfrac{50\sqrt{2}}{v} = 50\sqrt{2}[\dfrac{1}{v} +1] The shortest path outside the sand will be ARC. Time is taken to go from  A  to  C  via this path T_{outside} =  \dfrac{AR + RC}{1} =  \dfrac{25\sqrt{10} + 25\sqrt{10}}{1}              = 50\sqrt{10} \because   T_{sand}   1\sqrt{5} - 1 = 0.81ms^{-1}

Question 12

Starting from rest a particle moves in a straight line with acceleration

a = (25 - t^{2})^{1 / 2} m / s^{2} f o r 0 \leq t \leq 5 s

a = \frac{3 π}{8} m / s^{2} f o r t \geq 5 s

The velocity of particle at t = 7s is:

Accepted Answer

v=\int a\, dt a = (25 - t^2)^{1/2} m/s^2 \ for \ 0 \leq t \leq 5s      = \dfrac{3 \pi}{8} m/s^2 \ for \ t \geq 5s v =\int(25 - t^2)^{1/2} \,dt+a\ \  \  \  \ \ \ \ 0 \leq t \leq 5s      = \int\dfrac{3 \pi}{8} \,dt +b\ \ \ \ \ \ \ \ \ \ \  \ \ \ \ \  \ \ \ \ \ t \geq 5s v =\frac{t}{2}\sqrt{5^2-t^2}+\dfrac{25}{2}sin^{-1}(\dfrac{t}{5})+a\ \  \  \  \ \ \ \ 0 \leq t \leq 5s      = \dfrac{3 \pi t}{8}  +b\ \ \ \ \ \ \ \ \ \ \  \ \ \ \ \  \ \ \ \ \ t \geq 5s at  t=0 , v=0  gives  a=0 finding  b  using the fact that speed is a continous fuction lim_{t	o 5^{-}}  v=lim_{t	o 5^+}v 0+\dfrac{25}{2}sin^{-1}(\dfrac{5}{5})=\dfrac{3\pi 	imes 5}{8}+b b=\dfrac{35\pi}{8} v =\frac{t}{2}\sqrt{5^2-t^2}+\dfrac{25}{2}sin^{-1}(\dfrac{t}{5})\ \  \  \  \ \ \ \ 0 \leq t \leq 5s      = \dfrac{3 \pi t}{8}  +\dfrac{35\pi}{8}\ \ \ \ \ \ \ \ \ \ \  \ \ \ \ \  \ \ \ \ \ t \geq 5s so at  t=7    v= \dfrac{3 \pi 	imes 7}{8}  +\dfrac{35\pi}{8}\ \ \ \ \ \ \ \ \ \ \  \ \ \ \ \  \ \ \ \ \     v= \dfrac{8 \pi 	imes 7}{8} \ \ \ \ \ \ \ \ \ \ \  \ \ \ \ \  \ \ \ \ \  v=7\pi v=22m/s

Question 13

Question 14

A particle moves along a straight line and its velocity depends on time  t  as  v=4t-t^{2}.  Here  v  is in m/sec and  t  is in seconds. Then for the first  5  seconds:

Accepted Answer

Average velocity  =\frac{s}{\Delta t}=v_{avg} S=\int_{0}^{5}vdt=\int_{0}^{5}\left ( 4t-t^{2} ight )dt=\frac{25}{3}m v_{avg}=\dfrac{25/3m}{5 sec}=\dfrac{5}{3}\dfrac{m}{sec} Average speed  =\dfrac{dis tance\ covered}{time\ taken}=\dfrac{dis tance}{\Delta t} Distance  =\int_{0}^{4}v dt+\int_{4}^{5}\left ( -v ight )dt=\dfrac{32}{3}+\dfrac{7}{3}=\dfrac{39}{3}m =13 m Average speed  =\dfrac{13m}{5 sec} Average acceleration  \left ( a_{avg} ight )=\dfrac{v_{f}-v_{i}}{\Delta t} v_{f}=4 	imes 5-5^{2}=20-25=-5 v_{i}=0 a_{avg}=\dfrac{-5-0}{5}=-1 m/s^{2}

Question 15

The velocity-time graph of a body is shown in figure.

Accepted Answer

From  t=0  to  t=A  velocity increases from 0 to v where v is the velocity at  t=A From  t=A  to  t=B  velocity increases from v to 0 where v is the velocity at  t=A Hence,  v_{avg}  during interval  0 to A is same as  v_{avg}  during interval  A to B  \frac{v_c-0}{A-0} =	an 60^{\circ}=\sqrt{3} ;  	herefore v_c = \sqrt{3} 	imes OA \frac{v_c-0}{A-0} =	an 150^{\circ}=-	an 30^{\circ}=-\frac{1}{\sqrt{3}}  	herefore v_c = \frac{1}{\sqrt {3}} 	imes AB Equating both  v_c ,  \sqrt{3} 	imes OA = \frac{1}{\sqrt{3}} 	imes AB  \Rightarrow \frac{AB}{OA} =3 \Rightarrow  \frac{OA}{OA + AB} =\frac{1}{1+3}  \Rightarrow \frac{OA}{OB} =\frac{1}{4} Distance covered on in time OA:  d_{OA} = \frac{1}{2} 	imes OA 	imes v_A Distance covered on in time AB:  d_{AB} = \frac{1}{2} 	imes AB 	imes v_A |a_{OA} |=	an 60^{\circ}=\sqrt{3} |a_{AB}| =	an 30^{\circ}\frac{1}{\sqrt{3}} \frac{|a_{OA} |}{|a_{AB}| } =3  	herefore \frac{d_{OA}}{d_{AB}} = \frac{\frac{1}{2} 	imes OA 	imes v_A}{\frac{1}{2} 	imes AB 	imes v_A} 	herefore \frac{d_{OA}}{d_{AB}}=\frac{ \frac{AC}{\cot 60^{\circ}}}{\frac{AC}{\cot 30^{\circ}}}=	an 60^{\circ} 	imes \cot 30^{\circ}=3

Question 16

From point  A  located on a highway (shown in figure above) one has to get by car as soon as possible to point  B  located in the field at a distance  l  from the highway. It is known that the car moves in the field  \eta  times slower than on the highway. At a distance  \displaystyle CD=\frac{xl}{\sqrt{\eta^2-1}}  from point  D  one must turn off the highway. Find  x .

Accepted Answer

Let the car turn off the highway at a distance  x  from the point  D . So,  CD=x , and if the speed of the car in the field is  v , then the time taken by the car to cover the distance  AC=AD-x  on the highway \displaystyle t_1=\dfrac{AD-x}{\eta v}  (1)and the time taken to travel the distance  CB  in the field \displaystyle t_2=\dfrac{\sqrt{l^2+x^2}}{v}  (2)So, the total time elapsed to move the car from point  A  to  B \displaystyle t=t_1+t_2=\dfrac{AD-x}{\eta v}+\dfrac{\sqrt{l^2+x^2}}{v} For  t  to be minimum \displaystyle\dfrac{dt}{dx}=0  or  \displaystyle\dfrac{1}{v}\left[-\dfrac{1}{\eta}+\dfrac{x}{\sqrt{l^2+x^2}}\right]=0 or  \eta^2x^2=l^2+x^2  or  \displaystyle x=\dfrac{l}{\sqrt{\eta^2-1}} .

Question 17

A street car moves rectilinearly from station  A  (here car stops) to next station  B  (here also car stops) with an acceleration varying according to the law  f = a - bx , where  a  and  b  are positive constants and  x  is the distance from station  A . If the maximum distance between the two stations is  x = \dfrac {Na}{b}  then find  N .

Accepted Answer

Given that,&#10;&#10;The acceleration is  f=a-bx..... (I) &#10;&#10;The acceleration is decreasing with increasing x.&#10;&#10;Hence, the velocity will be maximum&#10;&#10;Where, f= 0,&#10;&#10;From equation (I),&#10;&#10;    a-bx=0   &#10;&#10;   x=\dfrac{a}{b}   &#10;&#10;Now, we know that&#10;&#10;    f=\dfrac{dv}{dt}  &#10;&#10;&#10;   f=\dfrac{dv}{dt}	imes&#10;\dfrac{dx}{dx}   &#10;&#10;   f=v\dfrac{dv}{dx}  &#10;&#10;&#10;Now, we can write,&#10;&#10;    v\dfrac{dv}{dx}=a-bx&#10;  &#10;&#10;  &#10;vdv=(a-bx)dx.....(II)   &#10;&#10;Now, for maximum velocity,&#10;&#10;   &#10;\int\limits_{0}^{{{v}_{\max }}}{vdv}=\int\limits_{0}^{\dfrac{a}{b}}{(a-bx)dx}  &#10;&#10;&#10;   \left[ \dfrac{{{v}^{2}}}{2}&#10;ight]_{0}^{{{v}_{\max }}}=\left( ax-b\dfrac{{{x}^{2}}}{2} ight)_{0}^{\dfrac{a}{b}}&#10;  &#10;&#10;   \dfrac{v_{\max&#10;}^{2}}{2}=\dfrac{{{a}^{2}}}{b}-\dfrac{b{{a}^{2}}}{2{{b}^{2}}}   &#10;&#10;   \dfrac{v_{\max&#10;}^{2}}{2}=\dfrac{2b{{a}^{2}}-b{{a}^{2}}}{2{{b}^{2}}}   &#10;&#10;   \dfrac{v_{\max&#10;}^{2}}{2}=\dfrac{{{a}^{2}}}{2b}   &#10;&#10;   v_{\max }^{2}=\dfrac{{{a}^{2}}}{b}&#10;  &#10;&#10;   {{v}_{\max }}=\dfrac{a}{\sqrt{b}}&#10;  &#10;&#10;Now, for the maximum distances x between the two stations&#10;&#10;On integrating of equation (II)&#10;&#10;At A and B, cars at rest so the velocity is zero at the&#10;both stations.&#10;&#10;Now,&#10;&#10;   &#10;\int\limits_{0}^{0}{vdv}=\int\limits_{0}^{x}{\left( a-bx ight)dx}   &#10;&#10;   0=\left( ax-b\dfrac{{{x}^{2}}}{2}&#10;ight)_{0}^{x}   &#10;&#10;   ax-b\dfrac{{{x}^{2}}}{2}=0&#10;  &#10;&#10;   a-b\dfrac{x}{2}=0  &#10;&#10;&#10;   x=\dfrac{2a}{b}......(III)&#10;  &#10;&#10;And given that,&#10;&#10; x=\dfrac{Na}{b}....(IV) &#10;&#10;Now, on comparing equation (III) and (IV)&#10;&#10;Then,&#10;&#10; N=2 Hence, the value of  N  is  2 .&#10;&#10;

Question 18

A particle is thrown vertically upward with a speed of 100 m/s . What is distance of particle travelled in 15 sec ?

Accepted Answer

herefore \ t=\dfrac {u}{g}=\dfrac {100}{10}=10\ sec \boxed {t=10\ sec} 	herefore \ v=0+10	imes 5 	herefore \ v=50\ m/sec Total distance  =x+y height, x   =\dfrac {u^2}{2g}=\dfrac {100	imes 100}{2	imes 10}=500\ m y=0(5)+\dfrac {1}{2}	imes 10 	imes 5^2 y=125\ m 	herefore \   total distance  =x+y =500+125 =625\ m

Question 19

A hunter is at  (4,-1,5)  units. He observes two preys at  P_{1}(-1,2,0)  units and  P_{2}(1,1,4)  respectively. At zero instant he starts moving in the plane of their positions with uniform speed of  5\: units \: s^{-1}  in a direction perpendicular to line  P_{1}P_{2}  till he sees  P_{1}  and  P_{2}  collinear at time  T . Time  T  is

Accepted Answer

Let,A be the vector acting along hunter and  P_1  andB be the vector acting along  P_1  and  P_2 Also,R be the vector of velocity with which hunter travels perpendicular to  P_1.P_2 Then, \vec {A}=(-1-4) \hat{i}+(2+1) \hat{j}+(0-5)\hat{k}=-5\hat{i}+3 \hat{j}- 5 \hat{k}  and \vec {B}=(1+1) \hat{i}+(1-2) \hat{j}+(4-0)\hat{k}=2 \hat{i}- \hat{j}+4 \hat{k}  Now,  R=|\vec {A}| \sin 	heta  and  |\vec {A}	imes\vec {B}|=AB \sin 	heta  or  \sin 	heta=\displaystyle \frac{|\vec{A} 	imes \vec{B}|}{AB} i.e  R=A \displaystyle \frac{|\vec{A} 	imes \vec{B}|}{AB}=\displaystyle \frac{|\vec{A} 	imes \vec{B}|}{B} Now,  \vec {A} 	imes \vec {B}=\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \  -5 & 3 & -5\ 2 & -1 & 4\end{vmatrix} =(12-5)\hat{i} + (-10+20)\hat{j} + (5-6) \hat{k}=7\hat{i}+10\hat{j}-\hat{k} |\vec {A} 	imes \vec {B}|=(7^{2}+10^{2}+1)^{1/2}=12.25 |\vec {B}|=(2^{2}+1^{2}+4^{2})^{1/2}=4.58 	herefore R=\displaystyle \frac{12.25}{4.58}=2.67 \:m Time taken to reach destination, =\displaystyle \frac{R}{v}=\displaystyle \frac{2.67}{5}=0.53 \:s

Question 20

A car leaves regularly from point  A  for point  B  every  10\ min . The distance between  A  and  B  is  60\ km . The car travel at a speed for  60\ km/h . Find the number of cars that a man driving from  B  to  A  will meet enroute if he starts from  B  simultaneously with one of the cars leaving  A . The car from B travels at a speed of  60\ km/h .

Accepted Answer

Time taken for car to travel from  B  to  A  is: T = \cfrac{s}{v} = \cfrac{60}{60} = 1\ hr = 60\ min Distance between any two consecutive cars leaving from A is constant and is given by: s  =10/60 	imes 60 = 10\ km Relative speed between any car leaving from  A  and car leaving from  B  is  v_r = 60 - (-60) = 120\ km/hr  Hence, on crossing one car, next car is met in time,  t = \cfrac{s}{v} = \cfrac{10}{120} \ hr= 5\ min Hence, total number of cars met are: n  =\cfrac{T}{t} = 12 This excludes the first car met at  B . Hence, car starting from  B  meets a total of  13\ cars .

Question 21

v-t  graph for rectilinear motion is given as shown in the figure. Matched the column on the basic of the  v-t  graph.

Accepted Answer

R.E.F image A) By definition : Avg. velocity =   \dfrac{total \, displacement}{total \, time \, taken} AS  \displaystyle V = \dfrac{dx}{dt} \Rightarrow 	riangle x = \int x.dt  So to find displacement, see the area under the from graph:  \Rightarrow  (Area under A)   m_{B}> M_{C}  \Rightarrow a_{A}> a_{B}> a_{C}    \boxed{Ans : 2,3} C) REF. Image.At   t = 0 : m_{C}> m_{B}> m_{A}  \Rightarrow a_{C} > a_{B} > a_{A}   \boxed{Ans :4} (D) Speed   \equiv   Magnitude of velocity.as velocity for all 3 answer has same value  t=t_{1} \Rightarrow \boxed{Ans :1}

Question 22

Study the following v-t graphs in column I carefully and match appropriately with the statements given in Column II. Assume that motion takes place from time 0 to T.

Accepted Answer

a. Area of v-t graph lies below the time axis, so displacement is negative, but slope is positive, so acceleration is positive.b. Area of v-t graph lies above the time axis, so displacement is positive, and slope is positive, so acceleration is also positive.c. Displacement is zero, because half area is above time axis and half below. Slope is negative, so acceleration is negative.d. Area of v-t graph lies above the time axis, so displacement is positive, and slop is negative, so acceleration is also negative.

Question 23

Question 24

when and where the ball will meet the elevator.

Accepted Answer

Let the ball and the elevator will meet after time  t  at the level where the ball has an upward displacement   x .For ball :     u  =18    m/s           a  = -g   = -9.8        m/s^2 using            S   = ut + \dfrac{1}{2}a t^2 	herefore          x  =  18 t  - \dfrac{9.8}{2} t^2 \implies        x  = 18 t - 4.9 t^2                              ..............(1)For elevator :         (7 + x  )   = v t 	herefore        7 + x  = 2t OR         7 + 18 t - 4.9 t^2  = 2t                                 [using (1)]OR          4.9 t^2  - 16 t  -7  = 0 \implies       t   = \dfrac{16 \pm \sqrt{16^2  + 4 (4.9) (7)}}{2 	imes 4.9}    = 3.65  s       OR          -0.39  s  (Not possible) 	herefore      x  = 2 (3.65)  -7   =0.3   mThus the ball and the elevator will at  12.30  m level  after   3.65   s

Motion In A Straight Line