A boat went 280 miles downstream and back.Going downstream took 7 hours Going back took 14. Whats speed of the boat in still water and the speed of the current?

Question

A math word problem that we are struggling with.

Can anyone help?

Answer 1

Assuming the boat traveled at a constant speed, and assuming the current is constant, and assuming that 280 miles is the length of the total trip rather then each way, we have:

 

140[miles] = 7[hours] x (v_boat + v_current)[miles/hour]

140[miles] = 14[hours] x (v_boat - v_current)[miles/hour]

 

Thus there are two independent equations and two variables which means that the system is solvable.

 

First solve for v_boat in the first equation:

 

v_boat = (140/7) - v_current

 

Then plug that into the second equation and solve for v_current:

 

140 = 14 x (140/7 - v_current -v_current)

 

(140/7) - (140/14) = 2 x v_current

v_current = 20/2 - 10/2 = 5 mph

 

Then plug that back into the earlier solution for v_boat:

 

v_boat = 140/7 - 5 = 15mph

 

Then plug the values into the original equations to double check:

 

140[miles] = 7[hours] x (15 + 5)[miles/hour] = 7 x 20 = 140 => good!
140[miles] = 14[hours] x (15 - 5)[miles/hour] = 14 x 10 = 140 => good!

so v_boat is 15mph and v_current is 5mph, but I would recommend getting clarification on whether the 280 miles was the total trip or if each leg of the journey was 280 miles for a total trip of 480.