Question

A fishing boat averages 10 mph in still water. It takes the boat 5 1/3 hr to travel 20 miles and return. Find the rate of the current.

Answer 1

vc = 14.43 mph

Explanation:

Given

vb = 10 mph (average speed of the fishing boat in still water)

t = 5 1/3 h = 16/3 h (total time)

d = 20 miles

if v = d/t ⇒ t = d/v

then

t₁ = 20 / (10+vc)

where 10+vc is the effective rate downstream

and

t₁ = 20 / (10-vc)

where 10-vc is the effective rate upstream

if we apply

t₁ + t₂ = t ⇒ (20 / (10+vc)) + (20 / (10-vc)) = 16/3

⇒ 2*vc² -15*vc - 200 = 0

Solving the equation we have

vc₁ = 14.43 mph and vc₂ = -6.93 mph

we choose vc = 14.43 mph