Case III:- River boat problems
Now initially a boat or swimmer is at one bank of river (let say at point A) swimmer or boat man wants to cross the river and to reach the other bank of river by the concept of relative motion.
Vbr = = Vb – Vr
Vb = Vbr + Vr
Along y-axis (perpendicular to river flow)
(Vb)y = (Vbr)y + (Vr)y
(Vb)y = Vbr + 0
Displacement along y-axis = d
Time taken to cross the river = t
Time take to cross the river
For minimum time or shortest time to cross the river
But in this case swimmer will not reach just opposite point he will cross the river and reach the other bank of river.
Drifting:- Horizontal distance due to river flow, when swimmer reach opposite bank of river. It is denoted by x
Now again Vbr = Vb - Vr
Along x –axis (Vbr)x = (Vb)x –(Vr)x
-Vbr sinq = (Vb)x –Vr
Disp. Along x-axis (Drifting) = x
Drifting in case of shorter time
Drifting = 0
Vr – Vbr sinq = 0
Unique angle to reach just opposite point of the river
Now General value of time
Down the stream and up the stream
Down the stream
up the stream
Note: Above Case is also applicable in escalator or moving belt.
Example: A river is flowing due east with speed 3 km/hr. A boat is crossing the river with speed 5 km/hr with respect to still water. Fnd time taken by the boat if boat reaches just opposite point on the other bank of river. Given river width is 0.5 Km.
Solution. Vr = 3km/h
Vbr = 5 km/h
d = 0.5 km
Reaching just opposite point
Case:- 4 Rain and umbrella problems
Case:- 4 Rain and umbrella problems
A person is moving with light walking speed of 3 km/h observe that rain is falling at angle of 37o with vertical if man increases his speed to 5 km/h he observe that rain is falling now in vertical direction. Find speed of rain with respect to ground.
Solution:-
Let velocity of rain is
Case I
3b = 4a-12
Case II:-
a = 5
Now 3b = 4a -12
3b = 4 x 5 - 12
Case I
A motorboat going downstream overcame a raft at a point A; t = 60 min later it turned back and after some time passed the raft at a distance l = 6.0 km below the point A. Find the velocity of the flow assuming that the engine worked in one regime in both directions .
Solution: Please see the video
Case 2
Two swimmers start swimming from point ‘A’ on one bank of the river to reach point ‘B’ lying right across on the other bank. One of them crosses the river along the straight line AB while the other swims at right angles to the stream and then walks the distances that the he has been carried away by the stream with velocity u to reach point ‘B’. Find at what value of u, both will reach point ‘B’ simultaneously? If the stream velocity is v0 = 2.0 km /h and velocity of each swimmer relative to the water is v’ = 2.5 km /h?
Solution: Please see the video
Case 3
An elevator car, whose floor – to –ceiling distance is 2.7m, starts ascending with constant acceleration a = 1.2 m/ s2; 2 s after the start a bolt starts falling from the ceiling of the car. Find the:
(a) bolt’s free fall time.
(b) Displacement and the distance covered by the bolt during the free fall in references frame fixed to the elevator shaft.
Solution: Please see the video
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Kaysons Academy
Course : JEE
Start Date : 04.06.2023
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Kaysons Academy
Course : JEE
Start Date : 03.06.2023
End Date : 30.06.2023
Types of Batch : Live Online Class