A golf ball is hit off a tee at the edge of a cliff.

Question

Its x and y coordinates as functions of time are given by x = 17.6t and y = 4.12t − 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 î and ĵ. (Give the answer in terms of t.)

By taking derivatives, do the following. (Give the answers in terms of t.)

(b) obtain the expression for the velocity vector v as a function of time

(c) obtain the expression for the acceleration vector a 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 = 2.85 s.

Answer 1

(a) Using the equations given for x and y, you can combine them into a vector that describes their position as a function of t.

Pos = 17.6t î +(4.12t - 4.90t²) ĵ

(b) To find the velocity vector, take the derivative of the position vector with respect to time.

Vel = 17.6 î +(4.12 - 9.80t) ĵ

(c) To find the acceleration vector, take the derivative of the velocity vector with respect to time.

Acc = 0 î - 9.80 ĵ

(d) Plug in t = 2.85 s into the Pos, Vel, and Acc vectors found in parts (a), (b), and (c)

Pos = 17.6(2.85) î +(4.12(2.85) - 4.90(2.85)²) ĵ = 50.16 î - 28.06 ĵ

Vel = 17.6 î +(4.12 - 9.80(2.85)) ĵ = 17.6 î - 23.81 ĵ

Acc = 0 î - 9.80 ĵ = - 9.80 ĵ