Animated Solution for Physics - Kinematics: A boy takes 60 min to swim across a river, if his goal is to minimize time; and takes 180 min, if his goal is to minimize to zero the distance that he is carried downstream. In both these attempts, the boy swims with the same speed relative to the river current. Which of the following statements can be true?
Select Answer:
* Multiple Correct
Visualized Solution
Visualizing the Two Strategies
Let v be the speed of the boy relative to water.
Let u be the speed of the river current.
Let d be the width of the river.
Case 1: Minimum Time
To minimize time, he must swim perpendicular to the river flow.
t_1 = \frac{d}{v} = 60 \text{ min} = 1 \text{ hr}
Case 2: Zero Drift
To have zero drift, he must swim upstream at an angle θ to the perpendicular.
His horizontal velocity must cancel the river current: vsinθ=u
x = \left(4 - 3\sqrt{2} \times \frac{1}{\sqrt{2}}\right) \times \sqrt{2}
x = (4 - 3)\sqrt{2} = \sqrt{2} \text{ km}
Since this is a possible scenario, Statement (d) can be true.
00:00 / 00:00
The Sigma Insight: Relative Velocity
Solution Diagram
The River Crossing Dilemma
Imagine you are standing on the bank of a steadily flowing river. You want to cross it, and you have two distinct goals in mind. First, you want to get to the other side as fast as humanly possible. Second, you want to cross without the river sweeping you downstream—you want to land exactly opposite your starting point.
Let's set up our mathematical stage. Let v be your swimming speed relative to the still water, u be the speed of the river current, and d be the width of the river.
Decoding Minimum Time
To cross the river in the absolute minimum time, you shouldn't waste any energy fighting the current. You must aim your body straight across to the opposite bank. The river will inevitably carry you downstream, but your crossing time depends entirely on your effort in the perpendicular direction.
So, your time t1 is simply the distance d divided by your speed v. The problem states this takes 60 minutes, or 1 hour:
t1=vd=1 hr
Mastering Zero Drift
Now, what if your goal is zero drift? To reach the exact opposite point, you have to fight the current by swimming at an angle θ upstream. Your horizontal velocity component, vsinθ, must perfectly cancel out the river's speed u.
Because you are angling yourself, your effective speed across the river is reduced to just vcosθ. This makes the journey longer. The problem states this takes 180 minutes, or 3 hours:
t2=vcosθd=3 hr
The Beautiful Cancellation
We have two powerful equations. Let's combine them by dividing the second time by the first time. Watch how the unknown distance d and speed v beautifully cancel out!
t1t2=vdvcosθd=cosθ1
Substituting our known times:
13=cosθ1⟹cosθ=31
Unveiling the Speeds
With cosθ known, we can easily find sinθ using the fundamental Pythagorean identity:
sinθ=1−cos2θ=1−(31)2=322
Now, remember our zero drift condition? The horizontal speeds must balance: u=vsinθ. Substituting our value for sinθ, we get:
u=v(322)
Since 322 is approximately 0.94, which is strictly less than 1, we can definitively say that v>u. Your swimming speed must be greater than the river's speed to achieve zero drift. Therefore, Statement (a) is true.
Testing the Scenarios
Let's put Statement (c) to the test. It proposes a specific river width: d=32 km.
Since you cross in 1 hour in the first case, your speed v must be exactly 32 km/h. Let's plug this v into our relation for u:
u=(32)(322)=4 km/h
The math aligns perfectly! The river speed is exactly 4 km/h. Thus, Statement (c) is true.
Finally, let's evaluate Statement (d). You now take 602 minutes, which is 2 hours. Let's assume you swim at a new angle α to the perpendicular. Your crossing time is:
t3=vcosαd⟹2=32cosα32
Solving this gives cosα=21, which means α=45∘.
The statement asks if you can be carried 2 km downstream. Let's assume you angle your 45∘ swim upstream. Your net horizontal speed is the river speed u minus your horizontal component vsin45∘:
Net Speed=4−32×21=4−3=1 km/h
Multiplying this net speed by the time 2 hours, your drift is exactly 2 km! Since this scenario is physically possible, Statement (d) can indeed be true.
