A golf ball is hit off a tee at the edge of a diff. Its x and y coordinates as f

Question

A golf ball is hit off a tee at the edge of a diff. Its x and y coordinates as functions of time are given by x = 19.0t and y = 3.68t - 4.90t^2, where x and y are in meters and t is in seconds. Write a vector expression for the ball's position as a function of time, using the unit vectors i cap and j cap. (Give the answer in terms of t.) r vector = m By taking derivatives, do the following. (Give the answers in terms of t.) obtain the expression for the velocity vector v vector as a function of time v vector = m/s obtain the expression for the acceleration vector a vector as a function of time a vector = m/s^2 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. r vector = m v vector = m/s a vector = m/s^2

Explanation / Answer

Ball' position will be x i^ + y j^.

Hence here s = 19t i^ + (3.68t -4.90t2) j^.

For velocity differentiate once w.r.t. time

V = 19i^+ (3.68 - 9.8t) j^.

For acceleration differentiate again

a = 0i^ - 9.8 j^.

Put t = 2.85 in each expression. a is constant for each time.

Dnt forget to assign beat answer.

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