Imagine a boat that travels at 3.0 m/s in still water. (a) In which direction should the boat point its bow to travel directly across a 0.50-km wide river when the river flows at 1.5 m/s? (b) What is the time interval for the trip, expressed in minutes?
This seems easy but I honestly have no idea how to do this. Please help.
Draw a vector diagram. Theta=arcsin(3/5). That will give the resulatant directly across.
time interval: speed across: 3*sinTheta
time= distance/speed= 500m/3sinTheta = ....seconds
a. Vr = Vb + Vc = Vb + (-1.5i) = 3.0 m/s = Resultant velocity.
Vb = 3 + 1.5i = 3.35m/s[26.6o]. N. of E.
b. d = V * T = 500 m.
3 * T = 500,
T = 166.667 s. = 2.78 min.
To answer this question, we need to understand the concept of vector addition. The boat's velocity can be broken down into two components: its velocity in still water and the velocity of the river. By vector addition, we can determine the resultant velocity of the boat relative to the riverbank.
(a) To find the direction the boat should point its bow, we need to determine the angle between the boat's path and the river's flow. Since the boat needs to travel directly across the river, we want the resultant velocity to be perpendicular to the river's flow.
Let's draw a diagram to visualize the problem:
```
----> |----->
Boat's Path |River's Flow
---->
Resultant Velocity
```
The boat's path and the river's flow are depicted as arrows, and we want the resultant velocity arrow to be perpendicular to the river's flow.
To find the angle between the river's flow and the boat's path, we can use trigonometry. Given that the boat's velocity in still water is 3.0 m/s and the river's flow velocity is 1.5 m/s, we have a right triangle with the following sides:
Opposite side: River's flow velocity = 1.5 m/s
Adjacent side: Boat's velocity in still water = 3.0 m/s
Using the formula tan(theta) = Opposite/Adjacent, we can solve for the angle theta:
tan(theta) = (1.5 m/s) / (3.0 m/s)
theta = tan^(-1)((1.5 m/s) / (3.0 m/s))
Calculating this on a calculator, we find that theta is approximately 26.56 degrees. Therefore, the boat should point its bow at an angle of 26.56 degrees upstream the river's flow to travel directly across the river.
(b) Now, let's determine the time interval for the trip, expressed in minutes. To do this, we need to calculate the boat's speed relative to the riverbank by finding the magnitude of the resultant velocity.
Using the Pythagorean theorem, we can find the magnitude of the resultant velocity:
Resultant velocity = sqrt((3.0 m/s)^2 + (1.5 m/s)^2)
Calculating this, we find that the magnitude of the resultant velocity is approximately 3.32 m/s.
Next, we need to calculate the time it takes for the boat to cross the 0.50-km wide river. Since distance = speed x time, we can rearrange the formula to solve for time:
Time = Distance / Speed
Time = (0.50 km) / (3.32 m/s)
To convert kilometers into meters, we need to multiply by 1000:
Time = (0.50 km * 1000 m/km) / (3.32 m/s)
Calculating this, we find that the time interval for the trip is approximately 150.6 seconds.
Finally, to express the time interval in minutes, we divide by 60:
Time = 150.6 seconds / 60 seconds/minute
Calculating this, we find that the time interval is approximately 2.51 minutes.
Therefore, the boat should point its bow at an angle of 26.56 degrees upstream, and the time interval for the trip is approximately 2.51 minutes.