Boat A east:
x = 13 t
Boat B north
y = 20 (t-1)
d = sqrt (x^2+y^2)
d = sqrt [169 t^2 + 400 (t^2-2 t +1) ]
d = (569 t^2 -800 t + 400 )^(1/2
Boat A leaves a dock headed due east at 1:00PM traveling at a speed of 13 mi/hr. At 2:00PM, Boat B leaves the same dock traveling due north at a speed of 20 mi/hr. Find an equation that represents the distance d in miles between the boats and any time t in hours for t≥1, using that t=0 corresponds to the time that Boat A leaves the dock.
1 answer
1 answer