Two boat landings are 1.0 km apart on the same bank of a stream that flows at 1.0 km/h. A motorboat makes the round trip between
the two landings in 50 minutes. What is the speed of the boat relative to the water?
1 answer:
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)
You might be interested in
Answer:
<em>The correct option is 1. 720 m</em>
Explanation:
<u>Projectile Motion</u>
When an object is launched in free air (no friction) with an initial speed vo at an angle , it describes a curve which has two components: one in the horizontal direction and the other in the vertical direction. The data provided gives us the initial conditions of the survival package's launch.
The initial velocity has these components in the x and y coordinates respectively:
And we know the plane has an altitude of 600 m, so the package will reach ground level when:
The vertical distance traveled is given by:
We'll set up an equation to find the time when the package lands
Solving for t, we find only one positive solution:
The horizontal distance is:
The correct option is 1. 720 m