Question

A dog goes through a park. Its displacement is described by the following equations:
x(t) = 28.0 + 7.2t -0.31t ² and y(t) = 30.0 -9.1t + 0.22t²

Positions and time are expressed in m and s, respectively.
a) At t = 15 s, determine the position vector r of the dog in terms of unit vectors i and .k
b) Determine for t = 15 s the velocity vector v of the dog terms of the unit vectors i and k.
c) Express this same velocity vector in polar coordinates, that is to say in terms of a module is a direction.

Answer 1

a] at t = 15s,

x = 28 + 7.2(15) - 0.31(15)² = 66.25m

y = 30 - 9.1(15) + 0.22(15)² = - 57m

so, the position vector r is: r = (66.25 m) i - (57 m) k

b] Differentiate individual expressions for x and y to get the velocity components along x and y.

v_x = dx/dt and v_y = dy/dt

v_x = 0 + 7.2 - 2(0.31)t

v_y = 0 - 9.1 + 2(0.22)t

at t = 15s,

v_x = - 2.1 m/s and v_y = - 2.5 m/s

the velocity vector is: v = - (2.1 m/s) i - (2.5 m/s) k

c] magnitude of velocity is:

$v = \sqrt{v_x^2+v_y^2} = 3.265 m/s$

and direction counter-clockwise from positive x axis is: $\theta = tan^{-1}\frac{v_y}{v_x} = 180 + 49.97 = 229.97\degree$ .