Q: A car moves with speed 60\ km/h for 1 hour in east direction and with same speed for 30\ min in south direction. The displacement of car from initial position is:

Given that, Speed v=60\,km/h Time {{t}_{1}}=1\,h {{t}_{2}}=30\,\min Now, distance travelled by a car in East direction {{D}_{1}}=v\times {{t}_{1}} {{D}_{1}}=60\times 1 {{D}_{1}}=60\,km Now, distance travelled by a car in south direction {{D}_{2}}=v\times {{t}_{2}} {{D}_{2}}=60\times \frac{1}{2} {{D}_{2}}=30\,km Now, the displacement from intial position D=\sqrt{{{\left( {{D}_{1}} \right)}^{2}}+{{\left( {{D}_{2}} \right)}^{2}}} D=\sqrt{{{\left( 60 \right)}^{2}}+{{\left( 30 \right)}^{2}}} D=\sqrt{3600+900} D=\sqrt{4500} D=30\sqrt{5} Hence, the displacement of car from initial position is 30\sqrt{5}\,m

Q: An automobile travelling with speed of 60\ {kmh}^{-1} can brake to stop within a distance of 20\ m . If the car is going twice, i.e., 120\ {kmh}^{-1} , the stopping distance will be

Third equation of motion : v^{2}=u^{2}+2as 0^{2}=(60\times(\dfrac{5}{18}))^{2}+2a\times 20 0=\dfrac{2500}{9}+40a a=-\dfrac{2500}{40\times 9} a=-\dfrac{125}{18} Now, when the initial speed is 120kmh^{-1} , acceleration remains same. v^{2}=u^{2}+2as 0^{2}=(120\times(\dfrac{5}{18}))^{2}+2s\times \dfrac {-125}{18} 0=\dfrac{10000}{9}-s\dfrac {125}{9} s=80 Therefore, the stopping distance will be 80 m

Q: A car moving on a straight road covers one third of a certain distance with 20 Km/h and the rest with 60 Km/h. The average speed is :

Given,Velocity for one third of distance {{V}_{1}}=20\,km/h Velocity for two third of distance {{V}_{2}}=60\,km/h Time take to cover one third of distance {{t}_{1}}=\dfrac{d/3}{{{V}_{1}}}=\dfrac{d/3}{20} Time taken to cover two third of distance {{t}_{2}}=\dfrac{2d/3}{{{V}_{2}}}=\dfrac{2d/3}{60} \text{Average velocity =}\dfrac{\text{Total}\,\text{distance}\,}{\text{total}\,\text{time}} {{V}_{av}}=\dfrac{d}{{{t}_{1}}+{{t}_{2}}}=\dfrac{d}{\dfrac{d/3}{20}+\dfrac{2d/3}{60}}=36\,km/h

Q: If a particle moving along a line following the law t=as^2+bs+c then the retardation of the particle is proportional to

Consider, t=as^2+bs+c Differentiate w.r.t. s \dfrac{dt}{ds}=2as+b Or, v=\dfrac{ds}{dt}=(2as+b)^{-1} ------- (i) Retardation=-a=\dfrac{-dv}{dt} Or, -a=\dfrac{dv}{ds}\times\dfrac{ds}{dt} -a=v\times\dfrac{dv}{ds} -a=v\times\frac{d}{ds}(2as+b)^{-1} -a=v(2as+b)^{-2}(2a) Putting the value of v from (i), -a=v(2as+b)^{-3}(2a) From this, it is evident that the retardation of the particle is proportional to the cube of the velocity. Option D.

Q: A river flows 3\ km/h and a man is capable of swimming at the rate of 2\ km/h . He wishes to cross it such that the displacement parallel to river is minimum. In which direction should he swim?

Let us assume he swims at an angle \theta with the perpendicular as shown in figure.If the river is l\,\,m wide, time taken to cross it, t=\displaystyle \frac{l}{2\cos \theta } , v_{x}=3-2 \sin \theta horizontal distance covered along x direction during this period x=v_{x}\times t . x=(3-2 \sin \theta)\displaystyle \frac{l}{2\cos \theta } for x to be minimum \displaystyle \frac{dx}{d\theta }=0 , or l [\displaystyle \frac{3}{2} \sec \theta \tan \theta-\sec ^{2} \theta]=0 or \sin \theta=\displaystyle \frac{2}{3}

Question 1

The shortest distance between the motorcyclist and the car is

Accepted Answer

Taking  N  as  +y-axis   and E as  +x -Axis  Imagine yourself as an observer siting inside the care You will regard the car as being at rest (at C) .relative to you the speed of the motorcyclist is obtained by imposing the reversed velocity of the car on motorcyclist as shown in fig  v_m=15ms^{-1},v_c=20ms^{-1} v_{mC}=\sqrt{15^2+20^2}=25,s^{-1} 	heta =	an ^{-1}\left(\dfrac{20}{15}ight)=53^o   with  X -axisThe motocyclist appears to moving the line  MP  with speed  25^{-1} The shortest distance =perpendicular distance of MP from  C=d ightarrow d=50 \cos53^o ightarrow d=30m

Question 2

A person goes to his office from his home daily at a fixed speed. But today he decided to reduce his speed by 10% as a result he reaches his office 20 minute late. The actual time he takes to reach his office daily is:

Accepted Answer

Let the speed by which he goes to his office be x.So, his speed by 20%, his final speed would be  80% of x =\dfrac{80x}{100} =  \dfrac{4x}{5} Initially:Speed =  \dfrac{d}{t} x = d/t -----------(1) xt = dAfter decrease in speed \dfrac{4x}{5} = \dfrac{d}{t}+ 16  -------(2)Or the equation can also be written as - 4xt + 64x = 5d Eliminating d from equation(1)&(2) 5xt = 5d   Rewriting Eq.1 4xt + 64x = 5d  ---- Eq.2 \dfrac{5xt}{4xt}+ 64x = 1 5xt = 4xt + 64x xt = 64x   t = 64  minutes

Question 3

A car moves with speed  60\ km/h  for  1  hour in east direction and with same speed for  30\ min  in south direction. The displacement of car from initial position is:

Accepted Answer

Given that,&#10;&#10;Speed v=60\,km/h  &#10;&#10;Time&#10;&#10;    {{t}_{1}}=1\,h  &#10;&#10;&#10;  &#10;{{t}_{2}}=30\,\min    &#10;&#10;Now, distance travelled by a car in East direction&#10;&#10;    {{D}_{1}}=v	imes&#10;{{t}_{1}}   &#10;&#10;   {{D}_{1}}=60	imes&#10;1   &#10;&#10;   {{D}_{1}}=60\,km  &#10;&#10;&#10;Now, distance travelled by a car in south direction&#10;&#10;    {{D}_{2}}=v	imes&#10;{{t}_{2}}   &#10;&#10;   {{D}_{2}}=60	imes&#10;\frac{1}{2}   &#10;&#10;   {{D}_{2}}=30\,km  &#10;&#10;&#10;Now, the displacement from intial position&#10;&#10;    D=\sqrt{{{\left(&#10;{{D}_{1}} ight)}^{2}}+{{\left( {{D}_{2}} ight)}^{2}}}   &#10;&#10;   D=\sqrt{{{\left(&#10;60 ight)}^{2}}+{{\left( 30 ight)}^{2}}}   &#10;&#10;   D=\sqrt{3600+900}  &#10;&#10;&#10;   D=\sqrt{4500}   &#10;&#10;   D=30\sqrt{5}   &#10;&#10;Hence, the displacement of car from initial position is  30\sqrt{5}\,m &#10;&#10;&#10;

Question 4

An automobile travelling with speed of  60\ {kmh}^{-1}  can brake to stop within a distance of  20\ m . If the car is going twice, i.e.,  120\ {kmh}^{-1} , the stopping distance will be

Accepted Answer

Third equation of motion : v^{2}=u^{2}+2as 0^{2}=(60	imes(\dfrac{5}{18}))^{2}+2a	imes 20  0=\dfrac{2500}{9}+40a a=-\dfrac{2500}{40	imes 9} a=-\dfrac{125}{18} Now, when the initial speed is  120kmh^{-1} , acceleration remains same. v^{2}=u^{2}+2as 0^{2}=(120	imes(\dfrac{5}{18}))^{2}+2s	imes \dfrac {-125}{18}  0=\dfrac{10000}{9}-s\dfrac {125}{9} s=80 Therefore, the stopping distance will be 80 m

Question 5

A car moving on a straight road covers one third of a certain distance with 20 Km/h and the rest with 60 Km/h. The average speed is :

Accepted Answer

Given,Velocity for&#10;one third of distance   {{V}_{1}}=20\,km/h   Velocity for&#10;two third of distance   {{V}_{2}}=60\,km/h    Time take to&#10;cover one third of distance   {{t}_{1}}=\dfrac{d/3}{{{V}_{1}}}=\dfrac{d/3}{20}&#10;  Time taken&#10;to cover two third of distance   {{t}_{2}}=\dfrac{2d/3}{{{V}_{2}}}=\dfrac{2d/3}{60}&#10;    	ext{Average&#10;velocity =}\dfrac{	ext{Total}\,	ext{distance}\,}{	ext{total}\,	ext{time}}  &#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;  {{V}_{av}}=\dfrac{d}{{{t}_{1}}+{{t}_{2}}}=\dfrac{d}{\dfrac{d/3}{20}+\dfrac{2d/3}{60}}=36\,km/h

Question 6

A particular moves vertically with an upward initial speed

v_{0} = 10.5 m s^{- 1}

. If its acceleration varies with time as shown in

a - t

graph in Fig. find the velocity of the particle at

t = 4 s

.

Accepted Answer

The velocity of the particle at  t=4\ s  can be given as \vec{v}_{4}=\vec{v}_{0}+\Delta \vec{v} where  \Delta \vec{v}\equiv A (= area under  a-t ) graph during first  4\ s )Referring to  a-t  graph (shown in the figure), we have  A=A_{1}+A_{2}-A_{3}-A_{4} .........(1)where  A_{1}=5	imes 1=5, \A_{2}=\dfrac{1}{2}	imes x	imes 5 A_{3}=\dfrac{1}{2}	imes (1-x)	imes 10    A_{4}=\dfrac{1}{2}	imes 2	imes 10=10 We can find the value of  x  as follows:Using properties of similar triangles, we have  \dfrac{x}{5}=\dfrac{1-x}{10} This yields  x=\dfrac{1}{3} Substituting  x=\dfrac{1}{3}  in  A_{2}=\dfrac{1}{2}	imes x	imes 5  and  A_{3}=\dfrac{1}{2}(1-x)	imes 10 , A_{2}=\dfrac{5}{6}  and  A_{3}=\dfrac{10}{3} Then substituting  A_{1}, A_{2}, A_{3}  and  A_{4}  in Eq. (1) we have A=-7.5\Rightarrow \Delta v=-7.5  and  \vec{v_{0}}=10.5\ m/s Hence, we have  v_{4}=v_{0}+\Delta v=10.5-7.5=3\ m/s

Question 7

If a particle moving along a line following the law  t=as^2+bs+c  then the retardation of the particle is proportional to

Accepted Answer

Consider,&#10;&#10; t=as^2+bs+c &#10;&#10;Differentiate w.r.t. s&#10;&#10; &#10;&#10; \dfrac{dt}{ds}=2as+b &#10;&#10;Or, &#10;&#10; v=\dfrac{ds}{dt}=(2as+b)^{-1} &#10;------- (i)&#10;&#10; &#10;&#10; Retardation=-a=\dfrac{-dv}{dt} &#10;&#10; &#10;&#10;Or,&#10; -a=\dfrac{dv}{ds}	imes\dfrac{ds}{dt} &#10;&#10; &#10;&#10; -a=v	imes\dfrac{dv}{ds} &#10;&#10; &#10;&#10; -a=v	imes\frac{d}{ds}(2as+b)^{-1} &#10;&#10; &#10;&#10; -a=v(2as+b)^{-2}(2a) &#10;&#10; &#10;&#10;Putting the value of&#10; v  from (i),&#10;&#10; &#10;&#10; -a=v(2as+b)^{-3}(2a) &#10;&#10; &#10;&#10;From this, it is&#10;evident that the retardation of the particle is proportional to the cube of the&#10;velocity.  Option D.

Question 8

A river flows  3\ km/h  and a man is capable of swimming at the rate of  2\ km/h . He wishes to cross it such that the displacement parallel to river is minimum. In which direction should he swim?

Accepted Answer

Let us assume he swims at an angle  	heta  with the perpendicular as shown in figure.If the river is  l\,\,m  wide, time taken to cross it, t=\displaystyle \frac{l}{2\cos 	heta } ,  v_{x}=3-2 \sin 	heta horizontal distance covered along  x  direction during this period  x=v_{x}	imes t . x=(3-2 \sin 	heta)\displaystyle \frac{l}{2\cos 	heta } for  x  to be minimum  \displaystyle \frac{dx}{d	heta }=0 , or  l [\displaystyle \frac{3}{2} \sec 	heta 	an 	heta-\sec ^{2} 	heta]=0  or \sin 	heta=\displaystyle \frac{2}{3}

Question 9

The position time graph of a moving particle is shown. Then which of the below best represents its velocity time graph :  (	heta_{1} = 	heta_{2} = 120 ^{0})

Accepted Answer

When displacement time curve is upward parabola then acceleration is constant and + ve. Therefore v-t graph for upward parabola is straight line of positive slope. Similarly, opposite for downward parabola.

Question 10

Two particles A and B start moving simultaneously along the line joining them in the same direction with acceleration of 1 m/s and 2 m/s and speeds 3 m/s and 1 m/s respectively. Initially A is 10 m behind B. What is the minimum distance between them?

A

4 m

B

8 m

C

12 m

D

16 m

View solution

Accepted Answer

Acceleration of A w.r.t B         a_{AB} = a_A - A_B  = 1 -2  =-1  m/s^2 Initial velocity of A w.r.t B           u_{AB} = u_A - u_B  = 3 -1  =2  m/s 	herefore  Distance traveled by A w.r.t B,         S_{AB}  =u_{AB} t + \dfrac{1}{2}a_{AB} t^2 OR         S_{AB}  = 2t - \dfrac{t^2}{2} For maximum distance traveled           \dfrac{dS_{AB}}{dt}  = 0   	herefore      2  - t  = 0               \implies  t =2  sThus maximum distance traveled        S_{AB}  = 2(2)  - \dfrac{2^2}{2}  = 2  m \implies  Minimum distance between the cars       D_{min}  = 10 - 2  =8  m

Motion In A Straight Line