1.A car is traveling a road that includes two sides of an equilateral triangle with a constant speed s. What is the magnitude of the average velocity v of the car?
2.A ship crosses a river aiming at the angle theta to the left from the straight courses. The speed of the ship with respect to water is v' . The width of the river is d. What is the water speed u ( positive direction to the right), if the side displacement of the ship as it lands on the other shore is h?
1. To find the magnitude of the average velocity of the car, we need to find the total displacement and divide it by the total time taken.
The car is traveling on two sides of an equilateral triangle, which means it covers equal distances for each side.
Let's assume the length of each side of the triangle is L. Therefore, the total distance covered by the car is 2L.
The time taken to cover each side of the triangle is given by t = L/s, where s is the constant speed of the car.
Since the car covers two sides, the total time taken is 2t = 2L/s.
Now, let's calculate the total displacement of the car. Since the car is back at the starting point after completing the triangle, the total displacement is zero.
Therefore, the magnitude of the average velocity (v) of the car is zero.
2. To find the water speed u, we can use the concept of vector addition.
Let V be the velocity of the ship relative to the ground, and W be the velocity of the water.
The horizontal component of V is given by Vx = v' * cos(theta), and the vertical component is Vy = v' * sin(theta).
Since the side displacement of the ship as it lands on the other shore is h, we can use the horizontal component of V to calculate the time it takes to cross the river.
The time can be calculated as t = d / Vx.
Now, let's consider the vertical component of V. Since the vertical displacement is zero (the ship lands on the other shore), we can equate the vertical displacement to zero:
0 = Vy * t - u * t,
where u is the water speed in the positive direction (to the right).
Substituting the values, we get:
0 = v' * sin(theta) * (d / Vx) - u * (d / Vx).
Simplifying the equation, we have:
0 = v' * sin(theta) - u.
Therefore, the water speed u is equal to v' * sin(theta).
To find the magnitude of the average velocity of the car and the water speed of the ship, we can use some basic physics concepts.
1. Magnitude of Average Velocity of the Car:
The average velocity is defined as the displacement divided by the time taken. In this case, the car is traveling along two sides of an equilateral triangle.
To find the displacement, we need to determine the total distance traveled by the car along the two sides of the triangle. Since it is an equilateral triangle, all sides have the same length. So, the total distance traveled is twice the length of one side of the triangle. Let's call this distance d.
Since the car is traveling at a constant speed s, the time taken to cover this distance is equal to d divided by s. The displacement is the length of the third side of the equilateral triangle, which is also equal to d.
Therefore, the average velocity v can be calculated as the displacement d divided by the time taken (d/s). Thus, the magnitude of the average velocity v of the car is d/s.
2. Water Speed of the Ship:
To find the water speed u of the ship, we can use the concept of vector addition. We know that the ship is moving in a straight course across the river aiming at an angle theta to the left.
Let's consider the following vectors:
- v' is the velocity of the ship with respect to the water (magnitude v', direction same as the ship's motion)
- u is the water speed of the ship (magnitude u, positive direction to the right)
- h is the side displacement of the ship as it lands on the other shore (magnitude h, positive direction to the right)
The side displacement h of the ship is equal to the horizontal component of the ship's velocity, which is v' * cos(theta). This is because the horizontal component represents the distance moved in the positive direction to the right (h) due to the water speed u.
So, we can write the following equation:
h = v' * cos(theta)
Now, to solve for u, we need to rearrange the equation:
u = h / cos(theta)
Therefore, the water speed u of the ship is equal to the side displacement h divided by the cosine of the angle theta.