A small child is learning to ride a bike for the first time. Her dad decides it would be a good idea to start her on a small hill. The hill is inclined at 3.00° to the horizontal. The child starts partway down the hill, 7.00 m away from her dad, who stays at the top. She starts to wobble, so her dad sprints toward her from rest with an acceleration of 3.70 m/s^2.

Question

How far has the child gone before her parent catches her?

Answer 1

the child's acceleration is ... g * sin(3º)

1/2 * 3.70 * t^2 = 1/2 * [9.81 sin(3º)] * t^2 + 7.00

solve for t , substitute back to find the child's travel distance

Answer 2

35