A golf ball is hit off a tee at the edge of a cliff. Its x and y coordinates as functions of time are given by x = 18.0t and y = 4.00t - 4.90t2, where x and y are in meters and t is in seconds. (a) Write a vector expression for the ball's position as a function of time, using the unit vectors and. By taking derivatives, obtain expressions for (b) the velocity vector as a function of time and (c) the acceleration vector as a function of time. (d) Next use unit-vector notation to write expressions for the position, the velocity, and the acceleration of the golf ball at t = 3.00 s.

Question

Mohammad Bond

Physics

14 July, 01:42

A golf ball is hit off a tee at the edge of a cliff. Its x and y coordinates as functions of time are given by x = 18.0t and y = 4.00t - 4.90t2, where x and y are in meters and t is in seconds. (a) Write a vector expression for the ball's position as a function of time, using the unit vectors and. By taking derivatives, obtain expressions for (b) the velocity vector as a function of time and (c) the acceleration vector as a function of time. (d) Next use unit-vector notation to write expressions for the position, the velocity, and the acceleration of the golf ball at t = 3.00 s.

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Eleanor · Answer 1

Position: x = 18t y = 4t - 4.9t²

First derivative: x' = 18 y' = 4 - 9.8t

Second derivative: x'' = 0 y'' = - 9.8

Position vector: P = (18t) i + (4t - 4.9t²) j

Velocity vector: V = (18) i + (4 - 9.8t) j

Acceleration vector A = ( - 9.8) j