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

Question

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 = 19.6t and y = 3.68t - 4.81t2,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 . (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 as a function of time
=

(c) obtain the expression for 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 = 2.70 s.
= m
= m/s
= m/s2

Explanation / Answer

a) You're given the x and y coordinates of the position, so to express the position vector as a function of time, you simply need to multiply the x and y coordinates by their respective unit vectors, i (for the x-coordinate) and j (for the y-coordinate), like so:

P(t) = (19.6t) i + (3.68t - 4.81 t2) j

b) The derivative of a vector-valued function is equal to the vector sum of the derivatives of its components, and the derivative of a position function is the velocity function. Therefore, simply take the derivative of each component in the position function with respect to time to find the velocity-vector function:

V(t) = (19.6) i + (3.68 - 9.62 t) j

c) The acceleration vector is the derivative of the velocity vector, which we just derived, so take the derivatives of each component of the velocity-vector function again to find the acceleration-vector function:

A(t) = (0) i - 9.62 j

d) Because we derived the expressions for the position, velocity, and acceleration - vectors above as functions of t, we simply need to substitute t = 2.70 s to find the respective associated vectors:

P(2.70) = (52.9) i + (-25.1) j

V(2.70) = (19.6) i + (-22.3) j

A(2.70) = (0) i - 9.62 j

Notice here that I rounded each component to the nearest 3 significant digits, because the given time was 2.70 s (with 3 sig. digs)

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