Answer:
a) Vx = -31.95 km/h b) Vy = -45.07 km/h
c) t = 0.083 h d) d = 0.22 km
Explanation:
First we have to express these values as vectors:
ra = (2.5, 3.9) km rb = (0,0)km
Va = (0, - 21) km/h Vb = (31.95, 24.07) km/h
Now we can calculate relative velocity:
[tex]V_{A/B} = V_{A} - V_{B} = (0, -21) - (31.95, 24.07) = (-31.95, -45.07) km/h[/tex]
For parts (c) and (d) we need the position of A relative to B and the module of the position will be de distance.
[tex]r_{A/B} = (2.5, 3.9) + (-31.95, -45.07) * t[/tex]
[tex]d = |r_{A/B}| = \sqrt{(2.5 -31.95*t)^{2}+(3.9-45.07*t)^{2}}[/tex]
In order to find out the minimum distance we have to derive and find t where it equals zero:
[tex]d' = \frac{-2*(2.5-31.95*t)*(-31.95)-2*(3.9-45.07*t)*(-45.07)}{2*\sqrt{(2.5 -31.95*t)^{2}+(3.9-45.07*t)^{2}}} =0[/tex]
Solving for t we find:
t = 0.083 h
Replacing this value into equation for d:
d = 0.22 km
