Q: On open ground, a motorist follows a track that turns to his left by an angle of 60^{\circ} after every 500 m . Starting from a given turn, the displacement of the motorist at the third turn is:

The path followed by the motorist is a regular hexagon ABCDEF of side length 500 m as shown in the figure. Let the motorist starts from A and takes the third turn at D. Therefore, the magnitude of this displacement is AD = AG +GD =500m + 500m =1000m

Q: Two particles A and B are moving with uniform velocity as shown in Fig. given at t = 0 .a. Will the two particles collide?b. Find out the shortest distance between two particles.

Observing A from the frame of B Let us draw \overset{\rightarrow}{v_{AB}} , \overset{\rightarrow}{v}_{AB} = \overset{\rightarrow}{v}_{A} -\overset{\rightarrow}{v}_{B} = \overset{\rightarrow}{v}_{A} + (-\overset{\rightarrow}{v}_{B}) tan \theta = \dfrac{10}{20} = \dfrac{1}{2} Again, tan \theta = \dfrac{AD}{CD} = \dfrac{AD}{40} = \dfrac{1}{2} \Rightarrow AD = 20 \Rightarrow DO = 10 \Rightarrow BC = 10 Hence, the shortest distance between the particles d_{short} = BC \cos \theta = 10 \cos \theta = \dfrac{10 \times 2}{\sqrt{5}} = 4 \sqrt{5} m Since the closest distance is non-zero, therefore, the particles will not collide.

Q: A person walks along a straight road from his house to a market 2.5\ km away with a speed of 5\ km/hr and instantly turns back and reaches the house with a speed of 7.5\ km/hr . The average speed of the person during the time interval 0 to 50 minutes is (in m/s ):

Distance between house and market d = 2.5 kmFrom House to market :Speed of the person v_1 = 5 km/hr \therefore Time taken t_1 = \dfrac{2.5}{5} hr = 0.5 hr = 30 minFrom market to house :Speed of the person v_2 = 7.5 km/hr \therefore Time taken t_2 = \dfrac{2.5}{7.5} hr = \dfrac{1}{3} hr = 20 minTotal distance covered D = 2.5 + 2.5 =5 km = 5000 mTotal time taken T = 30 + 20 = 50 min = 3000 s \therefore Average speed of the person v_{avg} = \dfrac{5000}{3000} =\dfrac{5}{3} m/s

Q: A particle covers 10 m in first 5s and 10 m in next 3 s . Assuming constant acceleration. Find initial speed, acceleration and distance covered in next 2 s .

It is given that a particle covers 10m in first 5s and 10m in next 3s. so using the equation of motion Case I s=ut+\dfrac{1}{2}a{{t}^{2}} 10=5u+\dfrac{1}{2}a{{(5)}^{2}} 20=10u+25a 4=2u+5a..............(1) Case 2 In next 3s the particle covers more 10m distance. So 20=8u+\dfrac{1}{2}a{{(8)}^{2}} 5=2u+8a.........(2) On solving equation (1) and (2) 4=2u+5a 5=2u+8a a=\dfrac{1}{3}\,m/{{s}^{2}} Put the value of a in equation (1) u=\dfrac{7}{6}m/s Now to find distance in next 10 s. total time will be 10s s=\dfrac{7}{6}\times 10+\dfrac{1}{2}\times \dfrac{1}{3}\times {{(10)}^{2}} s=28.33\,m Distance travelled in next 2 sec s=28.33-20=8.33\,m

Q: Two ships A and B are 10 km apart on a line running south to north. Ship A farther north is streaming west at 20 km/h and ship B is streaming north at 20 km/h. What is their distance of closest approach and how long do they to reach it?

Ships A and B are moving with same speed 20 km/h in the directions shown in figure. It is two dimensional, two body problem with zero acceleration.Let us find \displaystyle \vec{v}_{BA} \displaystyle \vec{v}_{BA}= \vec{v}_{B}-\vec{v}_{A} Here, \displaystyle \left | \vec{v}_{BA} \right |= \sqrt{\left ( 20 \right )^{2}+\left ( 20 \right )^{2}}= 20\sqrt{2}km/h i.e., \displaystyle \vec{v}_{BA}\:is\:20\sqrt{2} km/h at an angle of \displaystyle 45^{\circ} from east towards north. Thus, the given problem can be simplified as:A is at rest and B is moving with \displaystyle \vec{v}_{BA} in the direction shown in figure.Therefore, the minimum distance between the two is \displaystyle s_{min}= AC= AB\sin 45^{\circ} \displaystyle = 10\left ( \frac{1}{\sqrt{2}} \right )km \displaystyle = 5\sqrt{2}km And the desired time is: \displaystyle t= \frac{BC}{\left | \vec{v}_{BA} \right |}= \frac{5\sqrt{2}}{20\sqrt{2}}\:\left ( BC= AC= 5\sqrt{2}km \right ) \displaystyle = \frac{1}{4}h =15\ min

Question 1

The shortest distance a spider should crawl from one corner to the opposite corner of a cubical room of dimensions  3\; m  is

Accepted Answer

To understand the solution clearly, open the cubical box as shown in the figure.It is given in the question that the spider should crawl from one corner to the opposite corner of the room. The shortest distance from one corner to the other corner is the 3D diagonal. Let us consider one of the 3D diagonals  CH . Shortest distance  CH  is correct but  CH=\sqrt { { CA }^{ 2 }+{ AH }^{ 2 } } =3\sqrt { 2 }m   is not the correct answer for this question because this route is aerial and the spider can only crawl along the ground or walls. Route,  CH=CA+AH=(3\sqrt { 2 } +3)m  is also not correct as it is not the shortest.Shortest distance to crawl is  CH=\sqrt { { CG }^{ 2 }+{ GH }^{ 2 } } =3\sqrt { 5 }\ m

Question 2

A monkey standing on the ground wants to climb to the top of a vertical pole  13  m tall. He climbs  5  m in  1  s and then slips downwards  3  m in the next second. He again climbs  5  m in  1  s and slips by  3  m in the next second as so on. How much time will the monkey take to reach the top of the pole?

Accepted Answer

The monkey takes 1 second to go up  5\ m  and then, he comes down by  3\ m  in next one second. So, every  2\ sec  he moves up by  2\ m . To move up  8\ m , he will take  8\ sec . From  8\ m  point, he will take one second to move another  5\ m , i.e., height of  8+5 =   13\ m . So, he has reached the top. Hence, he reaches the top in  8s + 1s = 9s .

Question 3

Without any stoppage a person travels a certain distance at an average speed of 42 km/hr, and with stoppages he covers the same distance at an average speed of 28 km/hr. How many minutes per hour does he stop ?

Accepted Answer

Given that,Without any stoppage a person travels a certain distance at an average speed  v=42\ km/hr With any stoppage a person travels a certain distance at an average speed  v=28\ km/hr We know that,Difference in distance traveled in one hour D =41&#8722;28=14\ km Which is lost due to stoppages.So, Stoppages time is t= \dfrac{14}{42}\ hr Time in minutes is t=\dfrac{14}{42}	imes60 t= 20\ min He stop in  20\ min . Hence, Option D is correct

Question 4

Two identical trains take  3\ sec  to pass one another when going in the opposite direction but only  2.5\ sec  if the speed of one is increased by  50\ \% . The time one would take to pass the other when going in the same direction at their original speed is

Accepted Answer

Both identical trains have speed of  x and  y   m/s and sum of their lengths be  D Thus,  t=\dfrac{D}{(x+y)}=3 When speed of one is increased by 50%:  t=\dfrac{D}{(1.5x+y)}=2.5 3x+3y=3.75 x+2.5 y 0.5y=0.75x y=1.5x D=3(x+y)=3(2.5x)=7.5x Thus, time for train moving in same direction=  \dfrac{D}{(y-x)}=\dfrac{7.5x}{(1.5x-x)}=\dfrac{7.5}{0.5}=15 \ s

Question 5

The position of a particle along x-axis at time t is given by  x = 2 + t - 3t^2 . The displacement and the distance travelled in the interval  t = 0  to  t = 1  are respectively:

Accepted Answer

x=2+t-3{ t }^{ 2 }\longrightarrow 1 diff. w.r.t. time \cfrac { dx }{ dt } =1-6t\longrightarrow 2 again differentiate w.r.t. time \cfrac { d{ x }^{ 2 } }{ d{ t }^{ 2 } } =-6 Hence motion is retarding because \cfrac { { d }^{ 2 }x }{ d{ t }^{ 2 } } <0 Let at t time velocity of particle is zero \cfrac { dx }{ dt } =0=1-6t 	herefore t=\cfrac { 1 }{ 6 }  at  	herefore t=\cfrac { 1 }{ 6 }   velocity of particle is zero, then particle moves in back direction at  t=1sec Now, displacement =  final position  -  initial position =x(1)-x(0) =2+1-3-2 =-2m Distance =\left| x(1)-x\left( \cfrac { 1 }{ 6 }  ight)  ight| +\left| x\left( \cfrac { 1 }{ 6 }  ight) -x\left( 0 ight)  ight|  =\left| 2+1-3-\left( 2+\cfrac { 1 }{ 6 } -\cfrac { 3 }{ 36 }  ight)  ight| +\left| 2+\cfrac { 1 }{ 6 } -\cfrac { 3 }{ 36 } -2 ight|  =\left| -2-\cfrac { 1 }{ 6 } +\cfrac { 1 }{ 12 }  ight| +\left| 2+\cfrac { 1 }{ 6 } -\cfrac { 1 }{ 12 } -2 ight|  =\left| -2-\cfrac { 1 }{ 12 }  ight| +\left| \cfrac { 1 }{ 12 }  ight|  =2+\cfrac { 1 }{ 12 } +\cfrac { 1 }{ 12 }  =2+\cfrac { 1 }{ 6 } =\cfrac { 13 }{ 6 } =2.16m

Question 6

On open ground, a motorist follows a track that turns to his left by an angle of  60^{\circ}  after every  500 m . Starting from a given turn, the displacement of the motorist at the third turn is:

Accepted Answer

The path followed by the motorist is a regular hexagon  ABCDEF  of side length  500 m  as shown in the figure. Let the motorist starts from  A  and takes the third turn at  D. Therefore, the magnitude of this displacement is AD = AG +GD  =500m + 500m  =1000m

Question 7

Two particles  A  and  B  are moving with uniform velocity as shown in Fig. given at  t = 0 .a. Will the two particles collide?b. Find out the shortest distance between two particles.

Accepted Answer

Observing  A  from the frame of  B Let us draw  \overset{ightarrow}{v_{AB}} , \overset{ightarrow}{v}_{AB} = \overset{ightarrow}{v}_{A} -\overset{ightarrow}{v}_{B} = \overset{ightarrow}{v}_{A} + (-\overset{ightarrow}{v}_{B}) tan 	heta = \dfrac{10}{20} = \dfrac{1}{2} Again,  tan 	heta = \dfrac{AD}{CD} = \dfrac{AD}{40} = \dfrac{1}{2} \Rightarrow AD = 20 \Rightarrow DO = 10 \Rightarrow BC = 10 Hence, the shortest distance between the particles d_{short} = BC \cos 	heta = 10 \cos 	heta = \dfrac{10 	imes 2}{\sqrt{5}} = 4 \sqrt{5} m Since the closest distance is non-zero, therefore, the particles will not collide.

Question 8

A person walks along a straight road from his house to a market  2.5\ km  away with a speed of  5\ km/hr  and instantly turns back and reaches the house with a speed of  7.5\ km/hr . The average speed of the person during the time interval  0  to  50   minutes  is (in  m/s ):

Accepted Answer

Distance between house and market       d = 2.5  kmFrom House to market :Speed of the person       v_1 = 5  km/hr 	herefore  Time taken       t_1 = \dfrac{2.5}{5}    hr   = 0.5  hr   = 30  minFrom market to house :Speed of the person       v_2 = 7.5  km/hr 	herefore  Time taken       t_2 = \dfrac{2.5}{7.5}    hr   = \dfrac{1}{3}  hr   = 20  minTotal distance covered         D = 2.5 + 2.5  =5   km   = 5000  mTotal time taken            T = 30 + 20 = 50    min   = 3000  s 	herefore  Average speed of the person       v_{avg}  = \dfrac{5000}{3000}   =\dfrac{5}{3}    m/s

Question 9

A particle covers  10 m  in first  5s  and  10 m  in next  3 s . Assuming constant acceleration. Find initial speed, acceleration and distance covered in next  2 s .

Accepted Answer

It is given that a particle covers 10m in first 5s and 10m&#10;in next 3s. so using the equation of motion&#10;&#10;Case I&#10;&#10;  s=ut+\dfrac{1}{2}a{{t}^{2}}  &#10;&#10;  10=5u+\dfrac{1}{2}a{{(5)}^{2}}  &#10;&#10;  20=10u+25a  &#10;&#10;  4=2u+5a..............(1)  &#10;&#10;Case 2&#10;&#10;In next 3s the particle covers more 10m distance. So&#10;&#10;  20=8u+\dfrac{1}{2}a{{(8)}^{2}}  &#10;&#10;  5=2u+8a.........(2)  &#10;&#10;On solving equation (1) and (2)&#10;&#10;  4=2u+5a  &#10;&#10;  5=2u+8a  &#10;&#10;  a=\dfrac{1}{3}\,m/{{s}^{2}}  &#10;&#10;Put the value of a in equation (1)&#10;&#10; u=\dfrac{7}{6}m/s &#10;&#10;Now to find distance in next 10 s. total time will be 10s&#10;&#10;  s=\dfrac{7}{6}	imes 10+\dfrac{1}{2}	imes \dfrac{1}{3}	imes&#10;{{(10)}^{2}}  &#10;&#10;  s=28.33\,m  &#10;&#10;Distance travelled in next 2 sec&#10;&#10; s=28.33-20=8.33\,m

Question 10

Two ships A and B are 10 km apart on a line running south to north. Ship A farther north is streaming west at 20 km/h and ship B is streaming north at 20 km/h. What is their distance of closest approach and how long do they to reach it?

Accepted Answer

Ships A and B are moving with same speed 20 km/h in the directions shown in figure. It is  two dimensional, two body problem with zero acceleration.Let us find  \displaystyle \vec{v}_{BA} \displaystyle \vec{v}_{BA}= \vec{v}_{B}-\vec{v}_{A} Here,  \displaystyle \left | \vec{v}_{BA} \right |= \sqrt{\left ( 20 \right )^{2}+\left ( 20 \right )^{2}}= 20\sqrt{2}km/h i.e.,  \displaystyle \vec{v}_{BA}\:is\:20\sqrt{2}  km/h at an angle of  \displaystyle 45^{\circ}  from east towards north. Thus, the given problem can be simplified as:A is at rest and B is moving with  \displaystyle \vec{v}_{BA}  in the direction shown in figure.Therefore, the minimum distance between the two is \displaystyle s_{min}= AC= AB\sin 45^{\circ}                     \displaystyle = 10\left ( \frac{1}{\sqrt{2}} \right )km                     \displaystyle = 5\sqrt{2}km And the desired time is: \displaystyle t= \frac{BC}{\left | \vec{v}_{BA} \right |}= \frac{5\sqrt{2}}{20\sqrt{2}}\:\left ( BC= AC= 5\sqrt{2}km \right )     \displaystyle = \frac{1}{4}h     =15\ min

Motion In A Straight Line