A boat traveled 280 miles downstream and back. The trip downstream took 7 hours. Trip back took 14 hours. What is the speed of the boat in still water? What is the speed of the current?

Respuesta :

Assuming 280 miles is the total distance travelled:
Let b = boat speed in still water
Let c = current speed.
For the downstream trip the speed is b + c. In 7 hours at the speed of (b + c) mph the boat travels 140 miles.
7(b + c) = 140 .............(1)
For the upstream trip the speed is b - c. In 14 hours at the speed of (b - c) mph the boat travels 140 miles.
14(b - c) = 140 ............(2)
The left hand sides of equations (1) and (2) are equal. Therefore we can write
7b + 7c = 14b - 14c ...........(3)
Rearranging equation (3) we get
21c = 7b
c = b/3 .......................(4)
The value for c obtained in equation (4) should now be substituted into equation (1) which can then be solved to find the value of b.

Answer:

speed of the boat: 30 mi/h, speed of the current: 10 mi/h

Explanation:

- During the trip downstream, the boat traveled 280 miles with a speed of (v+c), where v is the speed of the boat and c is the speed of the current. Since the time taken is 7 h, we can write the following equation:

[tex]7(v+c) =280[/tex]

which is the equivalent of [tex]time \cdot speed = distance[/tex]

- During the trip upstream, the boat traveled 280 miles with a speed of (v-c), where v is the speed of the boat and c is the speed of the current. Since the time taken is 14 h, we can write the following equation:

[tex]14(v-c) =280[/tex]

So we have two equations that we can solve simultaneously to find v and c:

[tex]7(v+c)=280\\14(v-c)=280[/tex]

Solving:

[tex]7v+7c=280\\14v-14c=280[/tex]

Multiplying first equation by 2:

[tex]14v+14c=560\\14v-14c=280[/tex]

By adding 2nd equation to 1st one, we get

[tex]28v=840\\14v-14c=280[/tex]

From 1st equation we get

[tex]v=30[/tex]

Substituting into second one we get

[tex]14(30)-14c=280\\420-14c=280\\-14c=-140\\c=10[/tex]

So, we have:

- speed of the boat in stil water: 30 mi/h

- speed of the current: 10 mi/h