Algebra Word Problem

Question

A jet ski went 8 miles upstream in 1 hr. The return trip took 30 minutes . Assume that the speed of the current (and its direction) was constant during both parts of the trip. Find the speed of the boat and the speed of the current (water). 

 

Please solve using a system of linear equations with a full explanation. I need the answer by Monday.

Answer 1

b - c = 8mi/1 hr = 8 mph (boat slows down when going against current)

 

b + c = 8mi/.5 hr = 16 mph (boat speeds up when going with current)

 

rearrange top equation b = 8 mph + c

replace b of bottom equation 8 mph + c + c = 16 mph

2c = 8 mph

c = 4 mph (current)

b = 8 mph + 4 mph = 12 mph (boat)

check answer 12 + 4 = 16 (woohoo)

 

That was algebra answer.  The following uses simultaneous linear equations.

 

b - c = 8 mph

b + c = 16 mph

 

add two equations together

 

2b = 24 mph (note c and -c cancels out)

b = 12 mph (look above!  same answer!)

c = 16 - 12 = 4 mph

Answer 2

Hi Merth,

 

Let's try to finish this on Friday.

 

Let X = the speed of the jet ski in still water

Let Y = the speed of the current.

 

The distance upstream equals the distance downstream.

 

Distance = speed x time

 

Upstream--> 8 = (X - Y) x 1

 

Downstream--> 8 = (X + Y) x 1/2

 

Remember that 30 minutes equals 1/2 hour.  Since the upstream time is given in hours, the downstream time unit has to match.

 

Let's multiply the second equation by 2 to get rid of the fraction--> X + Y = 16

Add it to the first equation-->                                                       X - Y = 8

                                                                                                --------------

                                                                                                  2X = 24

Divide both sides by 2--> X = 12 mph = jet ski speed

Then the current speed = 4 mph

 

Plug these speeds back into the original problem to make sure that they work.

 

Have fun.

 

Dan