Relative Motion: Rain Example
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Rain is falling vertically with a speed of 12ms−1.A cyclist is moving east to west with a speed of
12√3ms−1.In order to protect himself from rain at what angle he should hold his umbrella w.r.t
vertical.
t = 0, it start dropping packets at constant time interval of I0. If R represents the separation between two consecutive points of impact on the ground then for the first 3 packetsR1:R2 is
- 1 : 1
- 2 : 1
- 1 : 2
- 1 : 3
A swimmer can swim in still water at $ 1.5 m{s}^{-1}$. The river is flowing with $ 2 m{s}^{-1}$ from left to right. The minimum time taken by swimmer to cross the river and the drift in that case are respectively
1 iver flow V=2ms d =200m
- 3 km/h
- 3√2 km/h
- 2√3 km/h
- 32 km/h
- 6 km/hr, inclined at an angle of 30∘ to the vertical towards the man's motion.
- 3 km/hr, inclined at an angle of 30∘ to the vertical towards the man's motion.
- 6 km/hr, inclined at an angle of 45∘ to the vertical towards the man's motion.
- 6 km/hr, inclined at an angle of 60∘ to the vertical towards the man's motion.
A flag is hoisted on the steamer boat.On the basis of the data given below find out the direction of flutter of
the flag
→vsteamer/water=10√3ms−1 towards North
→vwater=5ms−1 towards East
→vwind=5ms−1 towards East
(Assume the wind does not affect the steamer)
North
South
East
West
- √116 m/s
- √32 m/s
- √10 m/s
- 5 m/s
- 5√5 m/s, θ=tan−142
- 5√5 m/s, θ=tan−112
- 10 m/s, θ=tan−12
- 10 m/s, θ=tan−112
A river flows due south with a speed of 2.0 m/s. A man steers a motorboat across the river; his velocity relative to the water is 4 m/s due east. The river is 800 m wide. How far south of his starting point will be reach the bank?
800 m
200 m
500 m
400 m
- 3√3 km/h
- 3√2 km/h
- 3√5 km/h
- 5√3 km/h
- 40√3 m/s
- 11.5 m/s
- 20√3 m/s
- 17 m/s
- 1 km/h
- 3 km/h
- 4 km/h
- 5 km/h
- 5 km/h
- 10 km/h
- 10√2 km/h
- 20 km/h
- 1 km/h
- 3 km/h
- 4 km/h
- 5 km/h
- At an angle of tan−1(25) with the vertical towards the east
- At an angle of tan−1(25) with the vertical towards the west
- At an angle of tan−1(52) with the vertical towards the east
- At an angle of tan−1(52) with the vertical towards the west
- 20 m/s, 10 m/s
- 10 m/s, 20√3 m/s
- 10√3 m/s, 20 m/s
- 20 m/s, 10√3 m/s
- 5 km/h
- 10 km/h
- 10√2 km/h
- 20 km/h
- 6 km/hr, inclined at an angle of 30∘ to the vertical towards the man's motion.
- 3 km/hr, inclined at an angle of 30∘ to the vertical towards the man's motion.
- 6 km/hr, inclined at an angle of 45∘ to the vertical towards the man's motion.
- 6 km/hr, inclined at an angle of 60∘ to the vertical towards the man's motion.
- 30∘
- 45∘
- 37∘
- 53∘
- tan−1(3√35)
- tan−1(√37)
- tan−1(35)
- tan−1(3√37)
- 20 kmh−1
- 10√2 kmh−1
- 20√3 kmh−1
- 10 kmh−1
(Take east and vertically upward as positive x-axis and positive y-axis respectively.)
- (12√3^i+12^j) m/s
- (12√3^i+12√3^j) m/s
- (12^i−12√3^j) m/s
- (−12^i−12√3^j) m/s
- 4√7 ms−1
- 8√2 ms−1
- 7√3 ms−1
- 8 ms−1
- tan−1(0.4)
- tan−1(1)
- tan−1(√3)
- tan−1(2.6)
- 20 kmh−1
- 30 kmh−1
- 40 kmh−1
- 15 kmh−1
- 10√5 m/s
- 50√5 m/s
- 10√20 m/s
- 5√5 m/s
From the previous problem at what angle to the horizontal must be steered in order to reach a point on the opposite bank directly east from the starting point? (Assuming his speed w.r.t. to the river is 4 m/s)
0∘
30∘
45∘
60∘
A river flows due south with a speed of 2.0 m/s. A man steers a motorboat across the river; his velocity relative to the water is 4 m/s due east. The river is 800 m wide. How far south of his starting point will be reach the bank?
400 m
800 m
200 m
500 m
- 37∘ North of vertical
- 30∘ South of vertical
- 53∘ North of vertical
- 60∘ South of vertical