Q: A train is targeted to run from Delhi to Pune at an average speed of 80kph but due to repairs of track looses 2hrs in the first part of the journey. If then accelerates at a rate of 20kph2 till the speed reaches 100kph. Its speed is now maintained till the end of the journey. If the train now reaches station in time, find the distance from when it started accelerating?

If t is normal runing time thenHeret1=80x1+2t2=20100−80=1hrt3=100l−x1−x2Here x2=80(1)+21(20)(1)2=90 kmThus total time t=t1+t2+t3=80l(80x1+2)+1+(100l−x1−90)=80lSoving it gives l−x1=840 km

Q: A train of length 350 m starts moving rectilinearly with constant acceleration 3×10−2 m/s2;t=30 s after the start the locomotive headlight is switched on (event-1) and T=60 s after that event the tail signal light is switched on (event-2). Find the distance between these events in the reference frames fixed to the train and to the Earth. How and at what constant velocity V relative to the Earth must a certain reference frame F move for the two events to occur in it at the same point ?

Distance of event- 1 iss1=350+21×0.03×(30)2=363.5 mDistance of event- 2 is s2=21×0.03×(90)2=121.5 mDistance between event-1 and 2 frame of train=350 mDistance between event-1 and 2 frame of ground=s1−s2=242 mSpeed of frome in which both event occur at same placev=60242=4.03 m/s

Q: Two cars traveling towards each other on a straight road at velocity 10 m/s and 12 m/s respectively. When they are 150 m apart, both drivers apply their brakes and each car decelerates at 2 m/s2 until it stops. How far apart will they be when they have both come to a stop ?

Stopping distance of car-1=2×2(10)2=25 mStopping distance of car-2=2×2(12)2=36 mSeparation between cars=150−25−36=89 m

Q: Two bodies move in the same straight line at the same instant of time from the same origin. The first body moves with a constant velocity of 40 m/s and second starts with a constant acceleration of 4 m/s2. Find the time t that elapses before the second catches the first body. Find also the greatest distance between them prior to it and the time at which this occurs.

After time t40tt=21×4×t2=20 sDistance between then at time tx=40t−2t2x is max whendtdxxmax=40−4t=0⇒t=10 s=40(10)−2(10)2=200 m

Q: A small parachute dropped from a 30 m high cliff falls freely under gravity for 1.0 s and then attains a terminal velocity 1.2 m/s. 20.0 s later a stone is dropped from the cliff. Will the stone catch up with the parachute before it reaches the ground? (Take g=10 m/s2 )

Distance fallen by the parachute in first 1.0 s is=21gt2=21(10)(1)2=5 mRest of the height 30−5=25 m, it covers with uniform speed 1.2 m/s. Thus time taken by the parachute to reach the ground is =1.225=20.84 s.Total time taken by the parachute to reach the ground is =1+20.84 s=21.84 s.Stone is dropped after 20.0 s. Time taken to reach ground is =g2H=102×30=2.45 sTotal time taken by stone from the time of dropping the parachute is 20+2.45=22.45 sIt is clear that stone requires time longer than the parachute requires. Thus stone will not catch the parachute.

Question 1

A train is targeted to run from Delhi to Pune at an average speed of 80kph but due to repairs of track looses 2hrs in the first part of the journey. If then accelerates at a rate of 20kph2 till the speed reaches 100kph. Its speed is now maintained till the end of the journey. If the train now reaches station in time, find the distance from when it started accelerating?

Accepted Answer

If t is normal runing time thenHeret1​=80x1​​+2t2​=20100−80​=1hrt3​=100l−x1​−x2​​​Here x2​=80(1)+21​(20)(1)2=90 kmThus total time t=t1​+t2​+t3​=80l​(80x1​​+2)+1+(100l−x1​−90​)=80l​Soving it gives l−x1​=840 km

Question 2

A train of length 350 m starts moving rectilinearly with constant acceleration 3×10−2 m/s2;t=30 s after the start the locomotive headlight is switched on (event-1) and T=60 s after that event the tail signal light is switched on (event-2). Find the distance between these events in the reference frames fixed to the train and to the Earth. How and at what constant velocity V relative to the Earth must a certain reference frame F move for the two events to occur in it at the same point ?

Accepted Answer

Distance of event- 1 iss1​=350+21​×0.03×(30)2=363.5 mDistance of event- 2 is s2​=21​×0.03×(90)2=121.5 mDistance between event-1 and 2 frame of train=350 mDistance between event-1 and 2 frame of ground=s1​−s2​=242 mSpeed of frome in which both event occur at same placev​=60242​=4.03 m/s​

Question 3

Two cars traveling towards each other on a straight road at velocity 10 m/s and 12 m/s respectively. When they are 150 m apart, both drivers apply their brakes and each car decelerates at 2 m/s2 until it stops. How far apart will they be when they have both come to a stop ?

Accepted Answer

Stopping distance of car-1=2×2(10)2​=25 mStopping distance of car-2=2×2(12)2​=36 mSeparation between cars=150−25−36=89 m

Question 4

Two bodies move in the same straight line at the same instant of time from the same origin. The first body moves with a constant velocity of 40 m/s and second starts with a constant acceleration of 4 m/s2. Find the time t that elapses before the second catches the first body. Find also the greatest distance between them prior to it and the time at which this occurs.

Accepted Answer

After time t40tt​=21​×4×t2=20 s​Distance between then at time tx=40t−2t2x is max whendtdx​xmax​​=40−4t=0⇒t=10 s=40(10)−2(10)2=200 m​

Question 5

A car and a truck move in the same straight line at the same instant of time from the same point. The car moves with a constant velocity of 40 m/s and truck starts with a constant acceleration of 4 m/s2. Find the time t that elapses before the truck catches the car. Find also the greatest distance between them prior to it and the time at which this occurs.

Accepted Answer

After time t from start, the position of car from start iss1​=40tand the position of truck from start iss2​=21​ (4) t2When truck catches the car or  or ​s2​40tt​=s1​=2t2=20 s​Before this the distance between them iss=s1​−s2​s=40t−2t2This distance is maximum when dtds​=0, (using maxima-minima), which givesdtds​=40−4t=0ort=10 sAt t=10 s, distance between them will be maximum, which is given assmax​=40(10)−2(10)2=200 m

Question 6

A driver travelling at 30kph sees the light turn red at the intersection. If his reaction time is 0.6 s, and the car can decelerate at 4.5 m/s2, find the stopping distance of the car. What would the stopping distance be if the car were moving at 90kph.

Accepted Answer

Initial speed of car is =30kph=8.3 m/s.Reaction time of driver is 0.6 s, it is the duration she takes to put the brakes on and in this duration car travels with the uniform speed.The distance travelled by the car during her reaction time is =8.34×0.6=5 m.Now after travelling 5 m, car start decelerating at 4.5 m/s2. As final velocity of the car is zero, when it comes to rest, we have for its initial velocity u=2as​.The distance travelled by the car before coming to rest can be given as s=2au2​=2×4.5(8.34)2​=7.71 mTotal stopping distance is =5+7.71=12.71 m.If initial speed of car were 90kph or 25 m/s, the distance travelled by car during reaction time of driver will become 25×0.6=15 m and the second distance will change as s=2×4.5(25)2​=69.45 m.Total stopping distance is =15+69.45=84.45 m

Question 7

If a body travels half its total path in the last second of its fall from rest, find the time and height of its fall. Take g=10 m/s2

Accepted Answer

If H be the total height the body falls from rest, total time to fall is t=g2H​​It is given that the body falls H/2 distance in time (t−1), thus we have t−1=g2(H/2)​​ or from above equations, eliminating t, we haveg2H​​−gH​​=1or gH​​(2​−1)=1or gH​​=0.4141​orH=58 mTotal time of fall is t=g2H​​=10116​​=3.4 s

Question 8

A person sitting on the top of a tall building is dropping balls at regular intervals of one second. Find the positions of the 3rd ,4th  and 5th  ball when the 6th ball is being dropped. Take g=10 m/s2

Accepted Answer

When 6th  ball is being dropped, the positions of the other (previously fallen) balls can be calculated by using the time of falling of each ball till this instant.For 5th  ball, it was dropped just one second before. Thus it has fallen a distance =21​gt2=5 m.For 4th  ball, it was dropped two second before this instant. It has fallen a distance =21​(10)22=20 m.For 3rd  ball, it was dropped two second before this instant. Its depth is =21​(10)32=45 m.

Question 9

A small parachute dropped from a 30 m high cliff falls freely under gravity for 1.0 s and then attains a terminal velocity 1.2 m/s. 20.0 s later a stone is dropped from the cliff. Will the stone catch up with the parachute before it reaches the ground? (Take g=10 m/s2 )

Accepted Answer

Distance fallen by the parachute in first 1.0 s is=21​gt2=21​(10)(1)2=5 mRest of the height 30−5=25 m, it covers with uniform speed 1.2 m/s. Thus time taken by the parachute to reach the ground is =1.225​=20.84 s.Total time taken by the parachute to reach the ground is =1+20.84 s=21.84 s.Stone is dropped after 20.0 s. Time taken to reach ground is =g2H​​=102×30​​=2.45 sTotal time taken by stone from the time of dropping the parachute is 20+2.45=22.45 sIt is clear that stone requires time longer than the parachute requires. Thus stone will not catch the parachute.

Question 10

A balloon is going up with a uniform speed 20 m/s. It was at a height of 100 m from ground, when a stone is dropped from its basket. Find the time taken by the stone to reach the ground and the height of the balloon from the ground, when stone hits the ground. (Take g=10 m/s2 )

Accepted Answer

As balloon is going up, when stone is dropped, it will also has the same upward speed of 20 m/s. So it will first go up retarded by gravity and fall back towards ground. The practical situation is shown in figure-Here to find time we use second speed equation for free fall motionh=ut−21​gt2From figure, here we take u=20 m/s and h=−100 m.We have−100=20t−21​(10)t2Solving, it gives t=6.89 s or t=−2.89 sThus stone will take 6.89 s to reach the ground and it will follow the path shown in figure by dotted line.In this duration, balloon ascends a height=20×6.89=137.8 mHeight of balloon from ground at the instant, stone hitting the ground is =100+137.8=237.8 m.

Question 11

From the foot of a tower 90 m high, a stone is thrown up so as to reach the top of the tower. Two second later another stone is dropped from the top of the tower. Find when and where two stones meet. (Take g=10 m/s2 )

Accepted Answer

First stone is thrown so as to reach the top of the tower, so its initial velocity isu​=2gH​=2×10×90​=42.5 m/s​Let us take the time t=t0​, when the two stones meet at a height h above the foot of the tower. The first stone is travelled a height h in the duration t0​ and the second stone has fallen a distance (90−h) in time (t0​−2). From speed equations, we have For first stone h=42.5t0​−21​(10)t02​For second stone90−h=21​(10)(t0​−2)2Adding above two equation,22.5t0​t0​​=70=3.11 s​Thus height h is given ash​=42.5(3.11)−21​(10)(3.11)2=83.82 m​

Question 12

A girl is standing in an elevator that is moving upward at a velocity of 5 m/s and acceleration 2 m/s2, when she drops her handbag. If she was originally holding the bag at a height of 1.5 m above the elevator floor, how long will it take the bag to hit the floor.

Accepted Answer

Solution
In such problems, generally called elevator problems we can use the concept of relative acceleration. Here we solve the problem with respect to elevator or we assume that we are observing from inside of elevator. The concept of relative motion is discussed in detail in section 1.11 of this chapter. That much of detail is not required here.
Imagine the situation, if you are standing in an elevator accelerating up with an acceleration a. If you are holding a box in your hand, it is also accelerating up with the same acceleration. If you release it free, it falls with acceleration g towards the elevator floor, which is coming up with acceleration a. Here we can say that the approach acceleration of box towards the elevator floor is (g+a), and we assume that elevator floor is at rest and box is going down with respect to floor with this acceleration called relative acceleration. Similarly if elevator is going down, we take relative acceleration (g−a).
In this problem, elevator is going up with a velocity 5 m/s, and an acceleration 2 m/s2, when the bag is dropped. Bag's relative velocity with respect to elevator floor is zero as both at that instant have the same velocity but relative acceleration of the bag is taken as 10+2=12 m/s2.
The distance fallen by the bag is 1.5 m, as we are assuming elevator is at rest.
Thus time required is =a2H​​=122×1.5​​=0.5 s

Question 13

A truck starts from rest with an acceleration of 1.5 m/s2 while a car 150 m behind starts from rest with an acceleration of 2 m/s2. How long will it take before both the truck and car side by side, and how much distance is traveled by each?

Accepted Answer

If we take truck at rest, then with respect to truck the relative acceleration of car is 2−1.5=0.5 m/s2. Now car has to travel 150 m with initial velocity zero, hence it takes time =0.52×150​​=24.5sec.Distance traveled by car in 24.5sec is=21​at2=21​×2×(24.5)2=600 mDistance traveled by truck in 24.5sec is=600−150=450 m

Question 14

An elevator car whose floor to ceiling distance is equal to 2.7 m starts ascending with constant acceleration 1.2 m/s2. Two seconds after the start a bolt begins falling from the ceiling of the car. Find :(a) the time after which bolt hits the floor of the elevator.(b) the net displacement and distance traveled by the bolt, with respect to earth. (Take g=9.8 m/s2 )

Accepted Answer

(a) If we consider elevator at rest, then relative acceleration of the bolt isar​=9.8+1.2=11 m/s2Initial velocity of the bolt is 2.4 m/s and it is getting retarded with 11 m/s2. With respect to elevator initial velocity of bolt is 0 and it has to travel 2.7 m with 11 m/s2. Thus time taken can be directly given ast=a2s​​=112×2.7​​=0.7 seconds (b) Velocity of the elevator after 2sec is v=at=1.2×2=2.4 m/s upwards.In 0.7 seconds distance traveled by the elevator isl​=2.4×0.7+0.5×1.2×0.49=1.98 m​The final displacement of the bolt is 2.7−1.98=0.72 m.We can observe the actual motion of the bolt, it is shown in figure-1.8. It first goes up and then gets back down and hit the floor of the elevator at a distance 0.72 m below the starting point. To find the distance we find the distance it travels before coming to rest at the topmost point of its trajectory. At the time of detachment of the bolt, its velocity is 2.4 m/s.So its distance up to the top =2gu2​=19.6(2.4)2​=0.293 mTotal distance traveled by the bolt is =0.293×2+0.72=1.31 m

Question 15

A ball is allowed to slip from rest down a smooth incline plane, and the distances are marked every 2.0 s. If the second mark is made 1.6 m from the starting point, where are the first and fourth marks ?

Accepted Answer

Given that1.6a​=21​a(4)2 which gives =0.2 m/s2​First mark is ats1​=21​a(2)2=21​0.2×(2)2=0.4 mFourth mark is ats4​=21​a(8)2=21​×0.2×(8)2=6.4 m

Question 16

Water drops from the nozzle of a shower into the stall floor 176.4 m below. The drops fall at regular interval of time, the first drop striking the floor at the instant the fourth drop begins to fall. Find the location of the individual drops when a drop strikes the floor. Take g=9.8 m/s2.

Accepted Answer

Total time of fall is t=9.82×176.4​​=6 sTime internal between drops =2 sLocation of second drop =21​×9.8×(4)2=78.4 mLocation of third drop =21​×9.8×(2)2=19.6 m

Question 17

A particle is projected vertically upward from the ground at time t=0 and reaches a height h at t=T. Show that the greatest height of the particle is (gT2+2h)2/8gT2.

Accepted Answer

h=uT−gT2 givesu=Th+21​gT2​Maxhight inH=2gu2​=8gT2(2h+gT2)2​

Question 18

A circus artist maintains four balls in motion making each in turn to rise to a height of 5 m from his hand. Calculate the velocity with which he projects the balls. Where will the other three balls be at the instant when the fourth one is just leaving his hand ? Take g=10 m/s2.

Accepted Answer

Projection speed u=2gh​=10 m/sTotal time of motion t=g2u​=2 sInternal between balls =0.5 sHeight of 2nd  ball h2​=10(1.5)−21​(10)(1.5)2=3.75 mHeight of 3rd  ball h3​=10(1)−21​(10)(1)2=5 mHeight of 4th  ball h4​=10(0.5)−21​(10)(0.5)2=3.75 m

Question 19

A ball projected vertically upwards from A, the top of a tower reaches the ground in t1​ seconds. If it is projected vertically downwards from A with the same velocity, it reaches the ground in t2​ seconds. If it falls freely from A, show that it would reach the ground in t1​t2​​ seconds.

Accepted Answer

Given that−h=ut1​−21​gt12​…(1)h=ut2​+21​gt22​…(2)h=21​gt32​…(3)​Solving (1), (2) and (3) we gett3​=t1​t2​​

Question 20

A dog sees a flowerpot sail up and then back down past a window 5ft high. If the total time the pot is in sight is 1.0sec, find the height above the window that the pot rises. Take g=32ft/s2

Accepted Answer

If at the bottom and of window speed of potential is u and vv=u−32×0.5andv2=u2−2×32×5Solving we getu=16f/s;v=4f/secHeight above window the potential rises ish=2gv2​=2×32(u)2​=161​ft

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