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 5 18.0t and y 5 4.00t 2 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 i^ and j^. by taking derivatives, obtain expressions for (b) the velocity vector vs as a function of time and (c) the acceleration vector as 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 5 3.00 s.

Answer 1

Given positions are

`x = 18 t`

`y = 4 t - 4.9 t^2`

part a)

position vector is given as

`\vec r = x \hat i + y \hat j`

`\vec r = 18 t\hat i + (4 t - 4.9 t^2) \hat j`

part b)

velocity is given as

`v = (dr)/(dt)`

now by differentiation of above equation

`v = 18 \hat i + (4 - 9.8 t)\hat j`

Part c)

For the acceleration we can use

`a = (dv)/(dt)`

so here it is

`a = 0 \hat i - 9.8 \hat j`

Part d)

at t = 3 s

`\vec r = 54\hat i - 32.1 \hat j`

`\vec v = 18 \hat i - 25.4 \hat j`

`\vec a = - 9.8 \hat j`