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A ball A is dropped from a building of height 45 m. Simultaneously another identical ball B is thrown up with a speed 50 m s^{−1}. The relative speed of ball B w.r.t. ball A at any instant of time is (Take g = 10 m s^{−2}).
Here, u_{A }= −0 , u_{B }= +50ms^{−1}
a_{A }= −g , a_{B }= −g
uB_{A }= uB − u_{A }= 50ms^{−1 }− (0)ms^{−1 }= 50ms^{−1}
aB_{A }= a_{B }− a_{A} = −g − (−g) = 0
∵ v_{BA }= u_{BA }+ a_{BA}t(As a_{BA }= 0)
∴ v_{BA }= u_{BA}
As there is no acceleration of ball B w.r.t to ball A, therefore the relative speed of ball B w.r.t ball A at any instant of time remains constant (= 50ms^{−1}).
A ball A is thrown up vertically with a speed u and at the same instant another ball B is released from a height h. At time t, the speed of A relative to B is
Taking upwards motion of ball A for time t, its velocity is V_{A} = U  gt.
Taking downwards motion of ball B for time, its velocity is V_{B} = gt.
Relative velocity of A w.r.t. B
=V_{AB} = V_{A} (V_{B}) = (u  gt)  (gt) = u
Two cars A and B are running at velocities of 60 km h^{−1} and 45 km h^{−1}. What is the relative velocity of car A with respect to car B, if both are moving eastward?
Velocity of car A w.r.t. ground
∴ v_{AG }= 60 kmh^{−1}
Velocity of car B w.r.t. ground
∴ v_{BG} = 45 km h^{−1}
Relative velocity of car A w.r.t. B
v_{AB }= v_{AG }+ v_{GB}
=v_{AG }− v_{BG }= 15 km h^{−1 }(∵ v_{GB }= −v_{BG})
Two cars A and B are running at velocities of 60 km h^{−1} and 45 km h^{−1}. What is the relative velocity of car A with respect to car B, if car A is moving eastward and car B is moving westward?
Velocity of car A w.r.t. ground
∴ v_{AG }= 60 kmh^{−1}
Velocity of car B w.r.t. ground
∴v_{BG} = 45 km h^{−1}
Relative velocity of car A w.r.t. B
v_{AB }= v_{AG }+ v_{GB}
= v_{AG }− v_{BG }= 105 km h^{−1 }(∵ v_{GB }= −v_{BG})
On a twolane road, car A is travelling with a speed of 36 km h^{1}. Two cars B and C approach car A in opposite directions with a speed of 54 km h^{1} each. At a certain instant, when the distance AB is equal to AC, both being 1 km, B decides to overtake A before C does. The minimum required acceleration of car B to avoid an accident is
Velocity of car A ,
Velocity of car B ,
Velocity of car C ,
Relative velocity of car B w.r.t. car A
v_{BA }= v_{B}−v_{A }= 15ms^{−1}−10ms^{−1 }= 5ms^{−1}
Relative velocity of car C w.r.t. car A is
v_{CA }= v_{C}−v_{A }= −15ms^{−1 }− 10ms^{−1}=−25ms^{−1}
At a certain instant, both cars B and C are at the same distance from car A
i.e. AB − BC = 1km = 1000m
Time taken by car C to cover 1km to reach car A
In order to avoid an accident, the car B accelerates such that it overtakes car A in less than 40s. Let the minimum required acceleration be a. Then,
A bird is tossing (flying to and fro) between two cars moving towards each other on a straight road. One car has speed of 27 km h^{−1} while the other has the speed of 18 km h^{−1}. The bird starts moving from first car towards the other and is moving with the speed of 36 km h^{−1} when the two were separated by 36 km. The total distance covered by the bird is
Velocity of car A, v_{A }= +27kmh^{−1}
Velocity of car B, v_{B }= −18kmh^{−1}
Relative velocity of car A with respect to car B
=v_{A }− v_{B} = + 27kmh^{−1 }(−18kmh^{−1}) = 45kmh^{−1}
Time taken by the two cars to meet = 36 km/45 km h^{1 }= 0.8
Thus, distance covered by the bird
= 36kmh^{−1 }x 0.8h = 28.8km
A police van moving on a highway with a speed of 30km h^{−1 }fires a bullet at a thief's car speeding away in the same direction with a speed of 192km h^{−1}. If the muzzle speed of the bullet is 150ms^{−1} , with what relative speed does the bullet hit the thief's car?
Speed of police van w.r.t. ground
∴ v_{PG }= 30kmh^{−1}
Speed of thief’s car w.r.t. ground
∴ v_{TG }= 192kmh^{−1}
Speed of bullet w.r.t. police van
Speed with which the bullet will hit the thief’s car will be
v_{BT }= v_{BG }+ v_{GT }= v_{BP }+ v_{PG }+ v_{GT}
= 540kmh^{−1 }+ 30kmh^{−1 }− 192kmh^{−1}
(∵ v_{GT }= −v_{TG})
A bus is moving with a speed of 10ms^{−1} on a straight road. A scooterist wishes to overtake the bus in 100s. If the bus is at a distance of 1km from the scooterist with what speed should the scooterist chase the bus?
Let v_{s} be the velocity of the scooter, the distance between the scooter and the bus = 1000m,
The velocity of the bus = 10ms^{−1}
Time taken to overtake = 100s
Relative velocity of the scooter with respect to the bus = (v_{s }− 10)
1000/(vs − 10) = 100s
= vs = 20ms^{−1}
Two towns A and B are connected by a regular bus service with a bus leaving in either direction every T minutes. A man cycling with a speed of 20 kmh^{1} in the direction A to B notices that a bus goes past him every 18 min in the direction of his motion, and every 6 min in the opposite direction. The period T of the bus service is
Let vkmh^{−1} be the constant speed with which the bus travel ply between the towns A and B .
Relative velocity of the bus from A to B with respect to the cyclist = (v − 20)kmh^{−1}
Relative velocity of the bus from B to A with respect to the cyclist = (v + 20)kmh^{−1}
Distance travelled by the bus in time T (minutes) = vT
As per question
Equating (i) and (ii) , we get
= 18v − 18 x 20 = 6v + 20 × 6
or 12v = 20 x 6 + 18 x 20 = 480 or v = 40 kmh^{−1}
Putting this value of v in (i) , we get
40T = 18 x 40 − 18 x 20 = 18 x 20
A 175m long train is travelling along a straight track with a velocity 72km−^{1}h. A bird is flying parallel to the train in the opposite direction with a velocity 18km^{−1}h. The time taken by the bird to cross the train is
Velocity of train,
Since bird is flying parallel to train in opposite direction.
∴ Relative velocity of bird w.r.t. train = vB + vT = 25m/s Train’s length = 175m
Time taken by the bird to cross the train is = 175/25 = 7s
Two parallel rail tracks run northsouth. On one track train A moves north with a speed of 54kmh^{−1} and on the other track train B moves south with a speed of 90kmh^{−1}. The velocity of train A with respect to train B is
Let the positive direction of motion be from south to north. Velocity of train A with respect to ground
Velocity of train B with respect to ground
Relative velocity of train A with respect to train B is
Two parallel rail tracks run northsouth. On one track train A moves north with a speed of 54 kmh^{−1} and on the other track train B moves south with a speed of 90kmh^{−1}. What is the velocity of a monkey running on the roof of the train A against its motion with a velocity of 18 kmh^{1} with respect to the train A as observed by a man standing on the ground?
Let the velocity of the monkey with respect to ground be v_{M}G
Relative velocity of the monkey with respect to the train A
A train A which is 120m long is running with velocity 20m/s while train B which is 130m long is running in opposite direction with velocity 30m/s. What is the time taken by train B to cross the train A?
Length of train A = 120m
Velocity of train A, v_{A }= 20m/s
Length of train B = 130m
Velocity of train B, v_{B }= 30m/s
Relative velocity of B w.r.t. A = v_{B }+ v_{A }= 50m/s
(trains move in opposite directions)
Total path length to be covered by B = 130 + 120 = 250m
∴ Time taken by train B = 250 m/50m/s = 5 s
A jet airplane travelling at the speed of 500kmh^{−1} ejects its products of combustion at the speed of 1500kmh^{−1 }relative to the jet plane. The speed of the products of combustion with respect to an observer on the ground is
Veloity of jet plane w.r.t ground v_{jG} = 500 km h^{1}
Velocity of products of combustion w.r.t jet plane v_{CJ} = 1500 kmh^{1}
∴ Velocity of products of combustion w.r.t ground is v_{CG} = v_{CJ} + v_{JG} =  1500kmh^{1} + 500 kmh^{1}
= 1000 km h^{1}
ve sign shows that the direction of products of combustion is opposite to that of the plane
∴ Speed of the products of combustion w.r.t ground = 1000 km h^{1}
On a long horizontally moving belt, a child runs to and fro with a speed 9kmh^{−1} (with respect to the belt) between his father and mother located 50m apart on the moving belt. The belt moves with a speed of 4kmh^{−1}. For an observer on a stationary platform, the speed of the child running in the direction of motion of the belt is
Figure shows conditions of the question.
In this case,
Speed of belt w.r.t. ground
∴ v_{BG }= 4kmh^{−1}
Speed of child w.r.t. belt
∴ v_{CB} = 9kmh^{−1}
∴ For an observer on a stationary platform, speed of child running in the direction of motion of the belt is
v_{CG }= v_{CB }+ v_{BG }= 9kmh^{−1 }+ 4kmh^{−1}
= 13kmh^{−1}
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