physics: circular motion?

Child's Velocity, dC/dt = 0.25 r./sec.

Merry-Go-Round's Velocity = dM/dt

Merry-Go-Round's Acceleration, d²M/dt² = 0.01 r./sec.²

Child's Position = C(t)

Merry-Go-Round's Position = M(t)

C(t) = ∫ [dC/dt] dt

C(t) = ∫ [0.25) dt

C(t) = 0.25t + c

c = 0, so

C(t) = 0.25t

dM/dt = ∫ [d²M/dt²] dt

dM/dt = ∫ [0.01] dt

dM/dt = 0.01t + c

c= 0, so

dM/dt = 0.01t

M(t) = ∫ [dM/dt] dt

M(t) = ∫ 0.01t] dt

M(t) = (0.01t² / 2) + c

M(t) = 0.005t²

c = 0, so

M(t) = 0.005t²

The child and the horse meet when the distance between the two are zero, so

M(t) - C(t) = 0

0.005t² - 0.25t = 0

0.005t(t - 50) = 0

If the product of two factors equals zero, then one or both factors equal zero.

If 0.005t = 0,

t = 0

If t - 50 = 0,

t = 50

Since t = 0 is the start time,

It will take the child 50 secs. to reach the horse.

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A child, searching for his favorite fave wooden horse, is running on the ground around the edge of a stationary merry-go-around. The angular speed of the child has a constant value of 0.25 rad/s. At the instant the child spots the horse, one-quarter of a turn away, the merry-go-around starts to move (in the direction of the child is running) w/ a constant of 0.01 rad/s^2. What is the shortest time it takes for the child to catch up w/ the wooden horse?

one-quarter of a turn = ¼ * 2 * π = 0.5 * π This is initial angle between the child and horse.

The angle that the child must move to catch the horse = ω * t = 0.25 * t

The initial position of the horse = 0.5 * π ahead of the child.

The angle that the horse moves = ½ * α * t^2 = ½ * 0.01 * t^2

Final position of the horse = 0.5 * π + ½ * 0.01 * t^2

The final positions are equal.

0.5 * π + ½ * 0.01 * t^2 = 0.25 * t

0.5 * π + 0.005 * t^2 = 0.25 * t

Set up quadratic equation and solve for 2 values of t.

0.005 * t^2 – 0.25 * t + 1.57 = 0

Time = 7.365 seconds. This is the shortest time!

Or

Time = 42.635 seconds

Check:

0.5 * π + ½ * 0.01 * 7.365^2 = 1.84 rad

0.25 * 7.365 = 1.84 rad