A stream is 30. meters wide and its current flows southward at 1.5 meters per second. A toy boat is launched with a velocity of 2.0 meters per second eastward from the west bank of the stream
a. what is the magnitude of the boat's resultant velocity as it crosses the stream?
b. how much time is required for the boat to reach the opposite bank of the stream?
a. X = Hor = 2 m/s
Y = Ver = 1.5 vm/s,
tanA = Y/X = 1.5/2 = 0.75,
A = 36.9 Deg.
R = X / cosA = 2/cos36.9 = 2.5 m/s
= Resulting velocity.
b. d = 30 m / cos36.9 = 37.5 M,
d = v * t,
37.5 = 2.5t,
t = 37.5 / 2.5 = 15 s.
15
a. Well, it seems like our toy boat is going to have quite the adventure. To figure out the magnitude of its resultant velocity, we need to use some good old trigonometry.
Since the boat is traveling eastward and the stream is flowing southward, we have a classic case of a right triangle on our hands. The boat's velocity is the horizontal leg, and the stream's velocity is the vertical leg.
To find the magnitude of the resultant velocity, we can use the Pythagorean theorem: c² = a² + b². In this case, a is the boat's velocity (2.0 m/s) and b is the stream's velocity (1.5 m/s). Let's do the math!
c² = (2.0 m/s)² + (1.5 m/s)²
c² = 4.0 m²/s² + 2.25 m²/s²
c² = 6.25 m²/s²
Taking the square root of both sides, we find:
c = √6.25 m/s
So, the magnitude of the boat's resultant velocity as it crosses the stream is approximately 2.50 m/s. How's that for speed?
b. Now, let's figure out how much time our little toy boat needs to reach the opposite bank of the stream. Since the stream is 30.0 meters wide, we can use the formula time = distance / velocity.
The distance the boat needs to cross is the width of the stream, which is 30.0 meters. The boat's velocity is 2.0 m/s eastward.
time = 30.0 m / 2.0 m/s
time = 15.0 s
Therefore, our brave toy boat will take approximately 15.0 seconds to sail across the stream. Just be careful not to get seasick!
To solve this problem, we can use vector addition and the Pythagorean theorem.
a. To find the magnitude of the boat's resultant velocity as it crosses the stream, we need to find the resultant vector of the boat's eastward velocity and the stream's southward velocity.
Using the Pythagorean theorem, we can calculate the magnitude of the resultant velocity:
resultant velocity^2 = (eastward velocity)^2 + (southward velocity)^2
resultant velocity^2 = (2.0 m/s)^2 + (1.5 m/s)^2
resultant velocity^2 = 4.0 m^2/s^2 + 2.25 m^2/s^2
resultant velocity^2 = 6.25 m^2/s^2
Taking the square root of both sides, we get:
resultant velocity = √(6.25 m^2/s^2)
resultant velocity ≈ 2.5 m/s
So, the magnitude of the boat's resultant velocity as it crosses the stream is approximately 2.5 m/s.
b. To find the time required for the boat to reach the opposite bank of the stream, we can use the formula:
time = distance / velocity
The distance the boat needs to travel is the width of the stream, which is 30 meters.
The boat's velocity is its eastward velocity, which is 2.0 m/s.
Plugging the values into the formula, we get:
time = 30 m / 2.0 m/s
time ≈ 15 seconds
Therefore, it would take approximately 15 seconds for the boat to reach the opposite bank of the stream.
To answer these questions, we need to use vector addition and the concept of relative motion.
Let's break down the velocities involved:
1. Stream's velocity: The stream flows southward at 1.5 meters per second. We will represent it as a vector with magnitude 1.5 m/s pointing south.
2. Boat's velocity: The boat has a velocity of 2.0 meters per second eastward. We will represent it as a vector with magnitude 2.0 m/s pointing east.
a. To find the magnitude of the boat's resultant velocity as it crosses the stream, we need to find the vector sum of the boat's velocity and the stream's velocity. This can be done using vector addition.
1. Represent the stream's velocity as (-1.5 m/s, 0 m/s) since it is purely in the south direction.
2. Represent the boat's velocity as (2.0 m/s, 0 m/s) since it is purely in the east direction.
3. Add the two vectors component-wise: (2.0 m/s + 0 m/s, -1.5 m/s + 0 m/s) = (2.0 m/s, -1.5 m/s)
4. The resulting vector represents the boat's resultant velocity. Its magnitude can be found using the Pythagorean theorem:
magnitude = sqrt((2.0 m/s)^2 + (-1.5 m/s)^2) ≈ 2.5 m/s
Therefore, the magnitude of the boat's resultant velocity as it crosses the stream is approximately 2.5 m/s.
b. To find the time required for the boat to reach the opposite bank of the stream, we need to determine the distance it needs to travel.
1. The stream's width is given as 30 meters, and the boat's velocity is solely eastward. Therefore, the distance the boat needs to travel to cross the stream is also 30 meters.
2. To find the time, we need to divide the distance by the magnitude of the boat's resultant velocity:
time = distance / magnitude ≈ 30 m / 2.5 m/s ≈ 12 seconds
Therefore, it takes approximately 12 seconds for the boat to reach the opposite bank of the stream.