Question

A motorboat in still water travels at a rate of 45 miles per hour. The current today is traveling at a rate of 10 miles per hour. If it took an extra hour to travel upstream, how far was the trip one way?

Answer 1

If the boat can travel 45 mph in still water and the current is 10 mph, down stream, or with the current, has the boat moving at 55.  Upstream has the boat moving 35 mph.  It took t time to go with the current and an hour longer, t + 1, to go against it.  Going from point A to B and back to A again means that the trip there was the same distance as the trip back. So d is the same for both upstream and downstream.  The equation for upstream is d = 35(t + 1) or d = 35t + 35; the equation for downstream is d = 55t.  Since the distances are the same, the transitive property says that so are the rates and times; therefore, we can set those equal to one another.  35t + 35 = 55t.  Solve for t.  When we find that t = 7/4 (1 3/4 hours), we sub that in to solve for distance.  55(7/4) = 96 1/4 miles.

Answer 2

Answer:

One way trip was of 157.5 miles.

Step-by-step explanation:

A motorboat in still water travels at a rate of 45 miles per hour.

Velocity of current today = 10 miles per hour

Let distance 'd' is covered in 't' hours by the motorboat in still water.

So the first equation will be

45 = $\frac{d}{t}$

d = 45t -------------(1)

Now for the second equation

if motorboat travels upstream then the relative velocity of the motorboat will

be = velocity of motorboat in still water - velocity of current

= 45 - 10 = 35 miles/hour  

If motorboat travels upstream it takes extra one hour to cover the same distance.

So the equation will be

35 = $\frac{d}{(t+1)}$

d = 35(t + 1) --------------(2)

Now we equate the value of d in both the equations 1 and 2.

35(t + 1) = 45t

35t + 35 = 45t

45t - 35t = 35

10t = 35

t = $\frac{35}{10}$

t = 3.5 hours

Now we put t = 3.5 in equation 1

d =  45 × 3.5 = 157.5 miles

One way trip was of 157.5 miles.