Couple of relative velocity problems.

[Raheelp]

1.

A swimmer heads directly across a river, swimming at 1.7 m/s relative to the water. She arrives at a point 44 m downstream from the point directly across the river, which is 76 m wide.

(a) What is the speed of the river current?
1 m/s

(b) What is the swimmer's speed relative to the shore?
2 m/s

I have my picture all drawn out but I don't understand the material at all.

2. This one is easy but my answer keeps coming out wrong ?

The range of a cannonball fired horizontally from a cliff is equal to the height of the cliff. What is the direction of the velocity vector of the projectile as it strikes the ground? (Ignore any effects due to air resistance.)

3. Last one I couldn't do.

While walking between gates at an airport, you notice a child running along a moving walkway. Estimating that the child runs at a constant speed of 2.7 m/s relative to the surface of the walkway, you decide to try to determine the speed of the walkway itself. You watch the child run on the entire 25-m walkway in one direction, immediately turn around, and run back to his starting point. The entire trip takes a total elapsed time of 33 s. Given this information, what is the speed of the moving walkway relative to the airport terminal?

So brain dead when it comes to physics, sigh.

 

[tiny-tim]

Hi Raheelp !

1: Relative velocities are vectors, so they obey the laws of vector addition.

I assume you've drawn a vector triangle … what are the sides and angles of it?​

2: Show us your full calculations, and then we can see what went wrong, and we'll know how to help.

(3, we'll leave for the present)

 

[tomcue]

1. In the first problem, we can use the concept of relative velocity to solve for the speed of the river current and the swimmer's speed relative to the shore. Since the swimmer is swimming directly across the river, her velocity relative to the shore will be the combination of her velocity relative to the water and the velocity of the river current. We can use the following equation to solve for the speed of the river current:

Vswimmer = Vwater + Vcurrent

Substituting the given values, we get:

2 m/s = 1.7 m/s + Vcurrent

Solving for Vcurrent, we get Vcurrent = 0.3 m/s.

Therefore, the speed of the river current is 0.3 m/s and the swimmer's speed relative to the shore is 2 m/s.

2. In the second problem, we can use the concept of projectile motion to determine the direction of the velocity vector of the cannonball as it strikes the ground. Since the range of the cannonball is equal to the height of the cliff, we can use the following equation to solve for the initial velocity of the cannonball:

R = v^2 * sin(2θ) / g

Where R is the range, v is the initial velocity, θ is the angle of projection, and g is the acceleration due to gravity. Since we are ignoring air resistance, we can assume that the angle of projection is 45 degrees. Substituting the given values, we get:

h = v^2 * sin(90) / g

Solving for v, we get v = √(2gh).

Therefore, the direction of the velocity vector of the cannonball will be at an angle of 45 degrees with the horizontal as it strikes the ground.

3. In the third problem, we can use the concept of relative velocity and the given information to determine the speed of the moving walkway relative to the airport terminal. Since the child runs on the entire 25-m walkway in one direction and then back to his starting point in a total elapsed time of 33 seconds, we can set up the following equation:

Vwalkway = (2*distance)/(total elapsed time)

Substituting the given values, we get:

Vwalkway = (2*25 m) / 33 s = 1.5 m/s

Therefore, the speed of the moving walkway relative to the airport