Question

NAME SHOW YOUR WORKI 11. (18) A small stone is kicked from the edge of a cliff. Its x- and y-coordinates as meters and t is in seconds functions of time are given by x 17.2t and y -3.96t- 4,9or', where x and y are in a) Sketch a graph of horizontal position with respect to time and a graph of vertical position with respect to time. yim) im) b) Write a vector expression for the stone's position as a function of time (in m) using the unit vectors i and j. i + TE c) Obtain the expression for the velocity vector v as a function of time (m/s) in rectangular components. 0 d) Obtain the expression for the acceleration vector a as a function of time (m/s') in rectangular components.

Answer 1

a) The horizontal position of the stone is a linear function of time that goes through the origin and has a slope of 17,2:

-35 -30 -25 -20 15 1D 0.2 0.4 0. D.8 1.2 14 1.6 1.8 2.2 2.4 .6 2.8

b) The vertical position of the stone is a parabolic function of time, The zeros of the function are t=0 and

3, 96 m/sl /4,9 m/s0,808s

The maximum height is reached for t=0,404s (the vertex of the parabola):

f (t) 3,96 [m/s] 0,404s 4,9 m (0,404 0,8m

0.2 0.4 .6 0.8 12 1.4 18 2.2 2.4 2.6 2. .2 -2 -3 10

b) The expression for the position using vector notation is:

-( 17, 2t)it (3.96tー4, 9t2)j

c) The components of the velocity vector are the derivatives with respect to time of the components of the position vector:

i-lar(17, 2t) at(3.96t _ 4. 9t2)| j= (17, 2)it (3.96-9.8t)j

This is

vr- 17,2 m/s

Uy -3, 96 [m/s - 9,8

d) The components of the acceleration vector are the derivatives with respect to time of the components of the velocity vector:

$\vec{a}=\left [\frac{\mathrm{d} }{\mathrm{d} t}(17,2) \right ]\vec{i}+\left [\frac{\mathrm{d} }{\mathrm{d} t}(3,96-9,8t) \right ]\vec{j}=-9,8\vec{j}$

This is:

a -0 m/s

9.8 m s Cl