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.

Question

contestada

Respuesta :

aristocles aristocles · Answer 1 · 2018-09-19T03:49:43+02:00

Given positions are

[tex]x = 18 t[/tex]

[tex]y = 4 t - 4.9 t^2[/tex]

part a)

position vector is given as

[tex]\vec r = x \hat i + y \hat j[/tex]

[tex]\vec r = 18 t\hat i + (4 t - 4.9 t^2) \hat j[/tex]

part b)

velocity is given as

[tex] v = \frac{dr}{dt}[/tex]

now by differentiation of above equation

[tex] v = 18 \hat i + (4 - 9.8 t)\hat j[/tex]

Part c)

For the acceleration we can use

[tex]a = \frac{dv}{dt}[/tex]

so here it is

[tex]a = 0 \hat i - 9.8 \hat j[/tex]

Part d)

at t = 3 s

[tex]\vec r = 54\hat i - 32.1 \hat j[/tex]

[tex]\vec v = 18 \hat i - 25.4 \hat j[/tex]

[tex] \vec a = - 9.8 \hat j[/tex]