River Flow Velocity Calculation: Simple Vector Physics Problem

[ownedbyphysics]

Simple vector physics--please help!

I'm having a little trouble with this physics problem. It seems so simple but I am stumped.

A river flows with a uniform velocity v. A
person in a motorboat travels 0.942 km up-
stream, at which time a log is seen floating by.
The person continues to travel upstream for
51.3 min at the same speed and then returns
downstream to the starting point, where the
same log is seen again.
Find the flow velocity of the river. Assume
the speed of the boat with respect to the water
is constant throughout the entire trip. (Hint:
The time of travel of the boat after it meets
the log equals the time of travel of the log.)
Answer in units of m/s.

For this problem, I figured that the log traveled .942 km in the amount of time that the boat took to travel 51.3 min and back. I assumed that it was 53.1*2 min to get back to when the boat first saw the log, but I have no idea how to get the amount of time it took for the boat to go the last .942km. I'm not even sure if it's needed in this problem. Please help!

 

[jbsteele]

To solve this problem, you first need to calculate the time it took for the boat to travel 0.942 km upstream. The speed of the boat with respect to the water is constant, so we can use the equation distance = speed x time to calculate the time.Distance = 0.942 kmSpeed = v (the flow velocity of the river, which we are trying to find)Time = ?Therefore, 0.942 km = v x TimeSolving for time, we have Time = 0.942 km / v.Now that we have the time it took for the boat to travel 0.942 km, we need to calculate the total time it took for the boat to make the round trip (travel 0.942 km upstream, wait for the same log to pass again, and travel 0.942 km back downstream). This total time is equal to 2 x Time, where Time is the time it took for the boat to travel 0.942 km upstream.Total time = 2 x Time = 2 x (0.942 km / v)Finally, we can use the equation time = distance / speed to calculate the flow velocity of the river. We know the distance is 0.942 km and the time is the total time it took for the boat to make the round trip, which we just calculated.Time = Total time = 2 x (0.942 km / v)Distance = 0.942 kmSpeed = v (the flow velocity of the river, which we are trying to find)Therefore, 0.942 km = v x (2 x (0.942 km / v))Solving for v, we have v = 0.942 km / (2 x (0.942 km / v))Solving further, we have v = 0.942 km / 2(0.942 km)Finally, v = 0.5 km/s = 500 m/s.Therefore, the flow velocity of the river is 500 m/s.

 

[tomcue]

I can provide some guidance on how to approach this problem using vector physics principles. First, let's define some variables to help us solve the problem:

v = velocity of the river (m/s)
v_b = velocity of the boat (m/s)
d = distance traveled by the boat (m)
t = time taken by the boat to travel d (s)
t_l = time taken by the log to travel d (s)

Now, let's break down the problem into smaller parts.

1. The boat travels upstream for 0.942 km (942 m) at a constant velocity v_b for a time t. This can be represented by the equation d = v_b * t.

2. During this time, the log also travels the same distance d = v * t_l.

3. We are given that t = 51.3 min, which is equal to 3078 s.

4. Using the equation d = v_b * t, we can rearrange it to solve for v_b: v_b = d/t. Substituting in the values, we get v_b = 942/3078 = 0.306 m/s. This is the velocity of the boat with respect to the water.

5. Now, we need to find the velocity of the river, which is represented by v. We can use the fact that the boat and the log traveled the same distance d = v * t.

6. Substituting in the values, we get v = 942/3078 = 0.306 m/s. This is the velocity of the river.

7. Finally, we can convert the velocity into m/s by dividing by 60 (since 1 min = 60 s). This gives us a final answer of v = 0.306/60 = 0.0051 m/s.

I hope this helps you understand the problem better and how to solve it using vector physics principles. Remember to always define your variables and break down the problem into smaller parts to make it more manageable.