Answer:
a) y= y₀+vy*t , x= x₀+vx*t
b) they are closest at t= 29.23 s
c) r min = 63.79 ft
Step-by-step explanation:
a) denoting v as velocities and "₀" as initial conditions , then the position of Sven is given by the coordinate (0,y) where
y= y₀+vy*t
and the position of Rudyard is given by the coordinate (x,0) where
x= x₀+vx*t
b) the distance r between Sven and Rudyard is given by
r²=x²+y²
the distance will be minimum when the derivative of r with respect to the time is 0 . Then taking the derivative of the equation above
2*r*dr/dt = 2*x*dx/dt + 2*y*dy/dt
since dx/dt= vx and dy/dt= vy , then
r*dr/dt = x*vx+ y*vy
dr/dt = (x*vx+ y*vy)/r
assuming that r cannot be 0 , then
dr/dt =0 → x*vx+ y*vy = 0
(x₀+vx*t)*vx + (y₀+vy*t)*vy = 0
-(x₀*vx + y₀*vy) = (vx²+vy²)*t
t= -(x₀*vx + y₀*vy)/(vx²+vy²)
replacing values
t= -(x₀*vx + y₀*vy)/(vx²+vy²) = -[ 140 ft*(-6ft/s) + (-170 ft)*4 ft/s]/[ (-6ft/s)²+ (4 ft/s)²] = 29.23 s
then they are closest at t= 29.23 s
and the minimum distance will be
x = x₀ + vx*t = 140 ft+(-6ft/s)*29.23 s = -35.38 ft
y= y₀+vy*t = (-170 ft)+ 4 ft/s*29.23 s = -53.08 ft
r min = √(x²+y²)= 63.79 ft
r min = 63.79 ft
Note
to prove our assumption that r is not 0 , then x and y should be 0 at the same time. thus
0= y₀+vy*t → t = (-y₀)/vy = -140 ft/(-6ft/s) = 26.33 s
0= x₀+vx*t → t= (-x₀)/vx = -(-170 ft)/4 ft/s = 42.5 s
then r is never 0
. Hence:
