algebra downstream/ upstream problem I am having trouble with

Question

Brianna rows a boat downstream for 216 miles. The return trip upstream took 18 hours longer. If the current flows at 3 mph, how fast does Brianna row in still water?

Answer 1

v = speed in still water

 

w = current speed = 3 mph

 

t = downstream trip time

 

Downstream trip:  (v + w) t = 216 miles, w = 3

Upstream trip:  (v - w) (t + 18) = 216 miles, w = 3

 

You need to solve two equations in two unknowns v and t:

 

(i)   (v + 3) t = 216

(ii)  (v - 3) (t + 18) = 216

 

From (i) get t = 216 / (v + 3) and substitute it into (ii) and solve for v:

 

 (v - 3) [216 / (v + 3) + 18] = 216

 

216 (v - 3) / (v + 3) + 18(v - 3) = 216

 

Multiply through by (v + 3) and get

 

216 (v - 3) + 18 (v - 3) (v + 3) = 216 (v + 3)

 

216v - 648 + 18 (v^2 - 9) = 216v + 648

 

18 (v^2 - 9) = 1296

 

v^2 = 1296 / 18 + 9 = 81

 

v = √81 = 9 mph

 

t = 216 / (v + 3) = 216 / (9 + 3) = 18 hours

 

Test the solution:

 

Is (v + 3)t = 216?

 

(9 + 3)(18) = 216

 

Is (v - 3)(t + 18) = 216?

 

(9 - 3)(18 + 18) = 216

 

 

 

 

Answer 2

Hi Brooke!  

 

This can be set up as a system of two equations with two unknowns:  the speed Brianna rows in still water, v_b, (in mph) and the time of the trip downstream, t_u, in hours.

 

Assuming a constant velocity, the relationship between distance (d), speed (v), and time (t) is d = vt.

 

With the current helping her (downstream), Brianna rowed 216 miles in time t.  So, for this case:

 

216 = (v_b+3)t_u  (Brianna's still-water speed is aided by the current, making her overall speed faster)

 

Going back upstream the same distance, the velocity is working against her, so:

 

216 = (v_b-3)(t_u+18)

 

We can solve for either unknown using these equations.  The method I choose is substitution.

 

Rearranging the first equation above gives:  t_u = 216/(v_b+3).  We can substitute this for t_u in the second equation, giving:

 

216 = (v_b- 3)((216/(v_b+3))+18)

 

This will take some algebra to solve for v_b, but it can be done.

 

216 = 216((v_b-3)/(v_b+3))+18(v_b-3)  [simplify right side]

216(v_b+3) = 216(v_b-3) + 18(v_b-3)(v_b+3)  [multiply each term by (v_b+3) to get v_b out of the denominator]

216v_b + 648 = 216v_b - 648 + 18(v_b2 - 9) [simplify each side, including using difference of two squares rule on far right term]

1296 = 18v_b² - 162  [cancel out like terms on each side and simplify further]

18v_b² = 1296+162 = 1458

v_b² = 1458/18 = 81

v_b = 9 mph  [taking the positive square root]

 

To check this against the original equations, we must also solve for t_u:

 

t_u = 216/(v_b+3) = 216/12 = 18 hours

 

We know this satisfies the first equation, because we just used that to solve for t_u.  Let's check the second equation:

 

216 = (9-3)(18+18)? = 6*36? = 216 CHECK