Two children are playing on the seesaw. THe lighter child is 9 feet from the fulcrum, and the heavier child is 6 feet from the fulcrum. As the lighter child goes from the ground to the highest point, she travels through an angle of 35 degrees in 1/2 second.

a. find the angular velocity of each child.

b. What is the linear velocity of each child?

help pls

To find the angular velocity of each child, we can use the formula:

Angular velocity (ω) = Δθ / Δt

Where Δθ is the change in angle and Δt is the change in time.

a. Let's start by finding the angular velocity of the lighter child. We are given that the lighter child travels through an angle of 35 degrees in 1/2 second.

Angular velocity (ω1) = 35 degrees / (1/2) second
= 35 degrees * 2 seconds
= 70 degrees per second

So, the angular velocity of the lighter child is 70 degrees per second.

Now, let's calculate the angular velocity of the heavier child. Since the seesaw is a balanced system, the angular velocity of the heavier child will be the same as that of the lighter child.

Angular velocity (ω2) = 70 degrees per second

Therefore, the angular velocity of both children is 70 degrees per second.

b. To find the linear velocity of each child, we can use the formula:

Linear velocity (v) = r * ω

Where r is the distance of the child from the fulcrum and ω is the angular velocity.

For the lighter child:
Distance from fulcrum (r1) = 9 feet
Angular velocity (ω1) = 70 degrees per second

Therefore, the linear velocity (v1) of the lighter child is:
v1 = 9 feet * (70 degrees per second)
= 630 feet per second

For the heavier child:
Distance from fulcrum (r2) = 6 feet
Angular velocity (ω2) = 70 degrees per second

Therefore, the linear velocity (v2) of the heavier child is:
v2 = 6 feet * (70 degrees per second)
= 420 feet per second

So, the linear velocity of the lighter child is 630 feet per second, and the linear velocity of the heavier child is 420 feet per second.