Finding rate of the current

Question

A motorboat averages 20mph in still water. It takes the boat the same amount of time to travel 3miles with the current as it does to travel 2miles against the current. What is the current's rate? 

Answer 1

Given that the speed of the boat in still water = 20 mph

 

Let speed of current = x 

 

Effective Speed of Boat downstream(with current) = 20 + x

 

Effective speed of boat upstream (against current) = 20 - x

 

Time taken to travel is given by:

 

Distance / speed

 

Therefore,

Time taken to travel downstream = 3 /(20 + x)

Time taken to travel upstream = 2 / (20 - x)

 

Since both travel times are the same, we have:

 

3 /(20 + x) = 2 / (20 - x)

 

Taking cross products we have:

2(20 + x) = 3(20 - x) 

 

40 + 2x = 60 - 3x 

5x = 20 

x = 20/5

x = 4

 

Therefore the current's rate is 4 mph.

Answer 2

distance=rate*time

d=rt

3=(20+c)t

2=(20-c)t

3/(20+c)=t

2/(20-c)=t

equate the left sides

3/(20+c)=2/(20-c)

cross multiply

3(20-c)=2(20+c)

60-3c=40+2c

60-40=3c+2c

20=5c

c=20/5

c=4 mph is the rate of the current

Answer 3

Let the velocity of the current be V

 

resultant velocity of boat with the current = 20 + V

 

resultant velocity of boat against the current = 20 - V

 

time = distance/velocity

 

both times are equal;

 

hence; 3/(20+V) = 2/(20-V)

 

cross multiplying; 3(20-V) = 2(20+V); giving V = 4mph