A boat is crossing a river that is 200 meters wide. The boat's speed in still water is 5.0 m/s, and the river current flows at 3.0 m/s perpendicular to the boat's direction, as shown in the diagram.

What is the magnitude of the boat's resultant velocity relative to the ground?
[ Select ]
4.0 m/s
2.0 m/s
8.0 m/s
6.4 m/s
5.8 m/s

How long will it take for the boat to reach the opposite bank?
[ Select ]
50 s
The boat can never reach the other bank.
33 s
40 s
67 s

Question

A boat is crossing a river that is 200 meters wide. The boat's speed in still water is 5.0 m/s, and the river current flows at 3.0 m/s perpendicular to the boat's direction, as shown in the diagram.

What is the magnitude of the boat's resultant velocity relative to the ground? 
[ Select ]
4.0 m/s
2.0 m/s
8.0 m/s
6.4 m/s
5.8 m/s

How long will it take for the boat to reach the opposite bank? 
[ Select ]
50 s
The boat can never reach the other bank.
33 s
40 s
67 s

BlackTom AI · Accepted Answer

To tackle this problem, let's separate the motion into perpendicular components and relate them to the given measurements.

First, consider the boat's speed in still water: 5.0 m/s. The river current is 3.0 m/s and flows perpendicular to the boat's intended direction across the river. If the boat aims directly across, its motion relative to the ground is the vector sum of its velocity across the river (5.0 m/s) and the current along the river (3.0 m/s in the perpendicular direction).

Option 1 (magnitude of resultant velocity): The resultant speed relative to the ground is the magnitude of the vector sum, which by Pythagoras is sqrt((5.0)^2 + (3.0)^2) = sqrt(25 + 9) = sqrt(34) ≈ 5.83 m/s. Looking at the given choices, 5.8 m/s is the value that matches this magnitude most closely. Other options can be checked similarly: 4.0 m/s and 2.0 m/s are too small to combine with a 3.0 m/s current to reach 5.83 m/s, 8.0 m/s and 6.4 m/s do not correspond to the vector sum of 5.0 and 3.0 in perpendicular directions.

Option 2 (time to reach the opposite bank): The time to cross the width of the river depends on the boat's velocity component perpendicular to the banks, which is the 5.0 m/s component across the river (the current is along the river, not across it). The width is 200 m, so time t = distance / perpendicular speed = 200 m / 5.0 m/s = 40 s. Among the provided choices, 40 s matches this calculation. The other times correspond to incorrect interpretations: 50 s would imply a slower crossing component, 33 s or 67 s would require different crossing speeds or widths, and "The boat can never reach the other bank" is not applicable here.

Summary for the options:
- For magnitude: 5.8 m/s is the correct match to sqrt(34) ≈ 5.83 m/s; the other numbers do not represent the resultant speed from combining perpendicular components of 5.0 m/s and 3.0 m/s.
- For crossing time: 40 s is correct since the cross-river component is 5.0 m/s and the width is 200 m; the time would be 200/5 = 40 s. The remaining time options correspond to incorrect assumptions about the crossing velocity.

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