Motion in a Straight Line Test 12

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Motion in a Straight Line Test 12

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Question 1
1 / -0

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).

A
0

B
10 m s^-1

C
25 ms^-1

D
50 ms^-1

Solution

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_BAt(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).
Question 2
1 / -0

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

A
u

B
u - 2gt

C

D
2u

Solution

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
Question 3
1 / -0

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?

A
15 km h^-1

B
45 km h^-1

C
60 km h^-1

D
105 km h^-1

Solution

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)
Question 4
1 / -0

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?

A
15 km h^-1

B
45 km h^-1

C
60 km h^-1

D
105 km h^-1

Solution

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)
Question 5
1 / -0

On a two-lane 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

A
1 ms^-2

B
1.5 ms^-2

C
2 ms^-2

D
3 ms^-2

Solution

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,
Question 6
1 / -0

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

A
28.8 km

B
38.8 km

C
48.8 km

D
58.8 km

Solution

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⁻¹x 0.8h = 28.8km
Question 7
1 / -0

A police van moving on a highway with a speed of 30km h⁻¹fires 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?

A
95 ms^-1

B
105 m s^-1

C
115 ms^-1

D
125 m s^-1

Solution

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)
Question 8
1 / -0

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?

A
40 ms^-1

B
25 ms^-1

C
115 m s^-1

D
125 ms^-1

Solution

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
Question 9
1 / -0

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

A
4.5 min

B
9 min

C
12 min

D
24 min

Solution

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
Question 10
1 / -0

A 175m long train is travelling along a straight track with a velocity 72km−¹h. A bird is flying parallel to the train in the opposite direction with a velocity 18km⁻¹h. The time taken by the bird to cross the train is

A
35 s

B
27 s

C
11.6 s

D
7 s

Solution

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
Question 11
1 / -0

Two parallel rail tracks run north-south. 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

A
10 ms^-1

B
15 ms^-1

C
25 ms^-1

D
40 ms^-1

Solution

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
Question 12
1 / -0

Two parallel rail tracks run north-south. 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?

A
5 ms^-1

B
10 ms^-1

C
15 ms^-1

D
20 ms^-1

Solution

Let the velocity of the monkey with respect to ground be v_MG
Relative velocity of the monkey with respect to the train A
Question 13
1 / -0

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?

A
5 s

B
25 s

C
10 s

D
100 s

Solution

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
Question 14
1 / -0

A jet airplane travelling at the speed of 500kmh⁻¹ ejects its products of combustion at the speed of 1500kmh⁻¹relative to the jet plane. The speed of the products of combustion with respect to an observer on the ground is

A
500 km h^-1

B
1000 km h^-1

C
1500 km h^-1

D
2000 km h^-1

Solution

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
Question 15
1 / -0

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

A
4 km h^-1

B
5 km h^-1

C
9 km h^-1

D
13 km h^-1

Solution

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