Q: If the river is 300 meters wide and the boat travels directly across (perpendicular to the current), how long will it take to cross if its speed relative to the water is $$15$$ mph? (1 mph = 0.447 m/s)

$$\frac{300}{15 \times 0.447} = \frac{300}{6.705} \approx 44.7$$ seconds

Q: If the boat heads straight across the river, how far downstream will it be carried by the current by the time it reaches the opposite shore? (River width = 300 m, current = 15 mph, 1 mph = 0.447 m/s)

$$15 \times 0.447 \times 44.7 \approx 300$$ meters

Q: A rescue boat must reach a person directly across a 400 m wide river with a current of $$10$$ mph. If the boat's speed is $$20$$ mph, what is the minimum time to reach the person?

$$\frac{400}{20 \times 0.447} = \frac{400}{8.94} \approx 44.7$$ seconds

Question 1

A boat is crossing a river. The boat has a speed of 15 mph relative to the water, and the river current is 15 mph perpendicular to the boat's motion. What is the resultant speed of the boat relative to the shore?

Accepted Answer

$$\sqrt{15^2$ + 15^2$} = \sqrt{450} = 21.2$$ mph

Question 2

If the boat in the previous question wants to travel directly across the river (perpendicular to the current), what direction should it head relative to the current?

Accepted Answer

Upstream at an angle so its velocity cancels the current

Question 3

Given the boat's speed is $$15$$ mph and the river current is $$15$$ mph, what is the angle the boat must head upstream to move directly across the river?

Accepted Answer

$$45^\circ$$ upstream from straight across

Question 4

If the river is 300 meters wide and the boat travels directly across (perpendicular to the current), how long will it take to cross if its speed relative to the water is $$15$$ mph? (1 mph = 0.447 m/s)

Accepted Answer

$$\frac{300}{15 	imes 0.447} = \frac{300}{6.705} \approx 44.7$$ seconds

Question 5

If the boat heads straight across the river, how far downstream will it be carried by the current by the time it reaches the opposite shore? (River width = 300 m, current = 15 mph, 1 mph = 0.447 m/s)

Accepted Answer

$$15 	imes 0.447 	imes 44.7 \approx 300$$ meters

Question 6

Which vector addition method is most appropriate for determining the resultant velocity of the boat and river current?

Accepted Answer

Tip-to-tail method

Question 7

If the boat's speed increases to $$20$$ mph while the river current remains $$15$$ mph, what is the new resultant speed of the boat relative to the shore?

Accepted Answer

$$\sqrt{20^2$ + 15^2$} = \sqrt{400 + 225} = \sqrt{625} = 25$$ mph

Question 8

A swimmer can swim at $$2 m/s in $$still water. If the river current is $$1.5 m/s$$, what is the magnitude of the swimmer's velocity relative to the shore if she swims directly across?

Accepted Answer

$$\sqrt{2^2$ + 1.5^2$} = \sqrt{4 + 2.25} = \sqrt{6.25} = 2.5 m/s$$

Question 9

If the boat travels at $$15$$ mph and the river current is $$15$$ mph, what is the direction of the boat's resultant velocity relative to the shore?

Accepted Answer

$$45^\circ$$ downstream from straight across

Question 10

A rescue boat must reach a person directly across a 400 m wide river with a current of $$10$$ mph. If the boat's speed is $$20$$ mph, what is the minimum time to reach the person?

Accepted Answer

$$\frac{400}{20 	imes 0.447} = \frac{400}{8.94} \approx 44.7$$ seconds

Vector Addition in Linear Motion: Riverboat Problem

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