Relative Velocity in 2D
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A boat takes to travel and back in still water. If the velocity of water is r, the time taken for going upstream and coming back is:
Cannot be estimated with the information given
![](https://df0b18phdhzpx.cloudfront.net/ckeditor_assets/pictures/927458/original_Picture1.png)
- 5 km
- 5√2 km
- 2 km
- √3 km
- 30∘ with direction of river flow
- 60∘ with direction of river flow
- 120∘ with direction of river flow
- 150∘ with direction of river flow
![](https://s3-us-west-2.amazonaws.com/infinitestudent-images/ckeditor_assets/pictures/720384/original_21.png)
Their angles of projection are 30∘ and θ, respectively, with the horizontal. If they collide after time t in air, then which of the following is incorrect?
- θ=53∘ and they will have same speeds just before the collision.
- θ=53∘ and they will have different speeds just before the collision.
- x<(1280√3−960) m
- It is possible that the particles collide when both of them are at their highest point.
Three particles A, B and C are situated at the vertices of an equilateral triangle ABC of side 'd' at t = 0. Each of the particles moves with constant speed v. A always has its velocity along AB, B along BC and C along CA. At what time will the particles meet each other?
2d3v
2d√3v
d√3v
insufficient data
- 60∘
- 120∘
- 90∘
- 150∘
![](https://s3-us-west-2.amazonaws.com/infinitestudent-images/ckeditor_assets/pictures/727811/original_1.png)
- 0∘
- 45∘
- 60∘
- 90∘
- 1500 m
- 2400 m
- 1950 m
- 1000 m
- 15∘ west of north
- 60∘ west of north
- 45∘ west of north
- 30∘ west of north
- (sin−1115) East of North, 50 min
- (30∘+sin−1115) East of North, 50 min
- (60∘+sin−1115) East of North, 50 min
- (30∘+sin−1115) East of North, 60 min
√91=9.5, 12.85≃0.35
- 0.64 hr
- 1 hr
- 0.33 hr
- Not possible
- 30∘, 2.0 hr
- 75∘, 2532√2 hr
- 45∘, 2527√2 hr
- 45∘, 1532√3 hr
- 40√2 km/h N−E
- 40√2 km/h S−E
- 40√2 km/h N−W
- 40√2 km/h S−W
A bird is flying due east with a velocity of 4 m/s. The wind starts to blow with a velocity of 3 m/s due north. What is the magnitude of relative velocity of bird w.r.t. wind? Find out the angle it makes with the x-axis?
7m/s, tan−1(34)
7 m/s, tan−1(43)
5 m/s, tan−1(34)
5 m/s, tan−1(43)
Three particles A, B and C are situated at the vertices of an equilateral triangle ABC of side 'd' at t = 0. Each of the particles moves with constant speed v. A always has its velocity along AB, B along BC and C along CA. At what time will the particles meet each other?
2d3v
2d√3v
d√3v
insufficient data
- 2√5 at an angle tan−1(2) North of West.
- 2√5 at an angle tan−1(2) West of North.
- 4√2 at an angle tan−1(13) North of West.
- 4√2 at an angle tan−1(13) West of North.
- If he crosses the river in minimum time, x=duv
- x cannot be less than duv
- For x to be minimum, he has to swim in a direction making an angle of π2+sin–1(vu) with the direction of the flow of water.
- x will be maximum if he swim in a direction making an angle of π2−sin–1(vu) with the direction of the flow of water.
![](https://s3-us-west-2.amazonaws.com/infinitestudent-images/ckeditor_assets/pictures/727481/original_4.png)
- u√2, in the upstream at an angle 45∘ with the vertical
- √2u, in the upstream at an angle 45∘ with the vertical
- u√2, in the downstream at an angle 45∘ with the vertical
- √2u, in the downstream at an angle 45∘ with the vertical
![](https://s3-us-west-2.amazonaws.com/infinitestudent-images/ckeditor_assets/pictures/727506/original_6.png)
- 80∘ with the direction of flow of river
- 110∘ with the direction of flow of river
- 120∘ with the direction of flow of river
- 150∘ with the direction of flow of river
A boy throws a ball upwards with velocity v0 = 20 m/s . The wind imparts a horizontal acceleration of 4 m/s2 to the left.
The angle θ at which the ball must be thrown so that the ball returns to the boy's hand is ( g = 10m/s2 )
tan−1(12)
tan−1(0.2)
tan−1(2)
tan−1(0.4)
- 50 km h−1, 30∘ East of North
- 50 km h−1, 30∘ North of East
- 25 km h−1, 30∘ East of North
- 25 km h−1, 30∘ North of East
- 0.75 km
- 1.5 km
- 2 km
- 1 km
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 much time is required to cross the river?
187.6 s
400 s
173.91 s
200 s
- 60∘
- 90∘
- 120∘
- 150∘
Column IColumn IIi. Minimum distance for vmω>vωa.θ=sin−1(vmωvω)ii Minimum time for vmω≥vωb.−→vm⊥→vωiii. Minimum distance for vmω<vωc.−−→vmω⊥→vωiv. Minimum time for vmω<vωd.θ=sin−1vωvmω
- i- b, d ii- c iii- a iv- c
- i- b, d ii- a iii- a iv- c
- i- b, d ii- c iii- a iv- a
- i- b, c ii- c iii- a iv- c
![](https://df0b18phdhzpx.cloudfront.net/ckeditor_assets/pictures/1068208/original_original_02_S1.png)
- θ=12sin−12425
- θ=12sin−11225
- θ=12sin−11625
- θ=12sin−1925
- 50 km h−1, 30∘ East of North
- 50 km h−1, 30∘ North of East
- 25 km h−1, 30∘ East of North
- 25 km h−1, 30∘ North of East
- 10.5 m
- 25 m
- 20 m
- 35 m
![](https://s3-us-west-2.amazonaws.com/infinitestudent-images/ckeditor_assets/pictures/720374/original_8.png)
- 5 m
- 5√3 m
- 5√3 m
- 10√3 m
- 5√74 hr
- 4√35 hr
- 2√3 hr
- 3√24 hr