The speed of an object changes only when it is acted on by an unbalanced force.
1 answer:
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Refer to the figure shown below.
Let V = speed of the boat relative to the water
Given:
u = 1 km/h the speed of flowing water.
When traveling downstream from A to B, the actual speed of the boat is
V₁ = V + u = V + 1 km/h
When traveling upstream from B to A, the actual speed of the boat is
V₂ = V - u =V - 1 km/h
Because the distance Ab is 1 km, the time taken for the round trip is
t = (1 km)/(V+1 km/h) + (1 km)/(v-1 km/h)
= (V-1 + V+1)/(V² - 1)
= (2V)/(V² - 1)
The time for the round trip is 50 min = 5/6 h.
Therefore
(2V)/(V² - 1) = 5/6
5(V² - 1) = 12V
5V² - 12V - 5 = 0
Solve with the quadratic formula.
V = (1/10)*[12 +/- √(144 + 100)] = 2.762 or -0.362 km/h
Ignore negative speed, so that
V = 2.762 k/h
Answer:
The speed of the boat relative to the water is 2.76 km/h (nearest hundredth)
Let V = speed of the boat relative to the water
Given:
u = 1 km/h the speed of flowing water.
When traveling downstream from A to B, the actual speed of the boat is
V₁ = V + u = V + 1 km/h
When traveling upstream from B to A, the actual speed of the boat is
V₂ = V - u =V - 1 km/h
Because the distance Ab is 1 km, the time taken for the round trip is
t = (1 km)/(V+1 km/h) + (1 km)/(v-1 km/h)
= (V-1 + V+1)/(V² - 1)
= (2V)/(V² - 1)
The time for the round trip is 50 min = 5/6 h.
Therefore
(2V)/(V² - 1) = 5/6
5(V² - 1) = 12V
5V² - 12V - 5 = 0
Solve with the quadratic formula.
V = (1/10)*[12 +/- √(144 + 100)] = 2.762 or -0.362 km/h
Ignore negative speed, so that
V = 2.762 k/h
Answer:
The speed of the boat relative to the water is 2.76 km/h (nearest hundredth)