Kayak Race

Question

In a kayak race, the participants must paddle a kayak 450 meters down a river & then return 450 meters up the river to the starting point. Susan has correctly deduced that the total time t(in sec) depends on the speed c(in m per sec) of the eater according to the following formula. Please fill in the chart.


Formula: t=450/v+c + 450/v-c
Time(t)sec v(m/sec) c(m/sec)
240          ?                1

300          ?                2
?              4                3
?              3                1

540          3                ?
?              3                3

Answer 1

The first two lines ask us to solve for v. We can substitute what we know first, or start algebraically:

 

240 = 450/(v+1) + 450/(v-1)

8/15 = 1/(v+1) + 1/(v-1)                      (dividing by 30, then by 15)

(8/15) (v+1) (v-1) = (v-1) + (v+1)       (multiplying by both denominators)

(8/15) (v²-1) = 2v

4v² - 4 = 15v                                       (multiplying by 15 and dividing by 2)

4v² - 15v - 4 = 0

 

t = 450/(v+c) + 450/(v-c)
(t/450) = 1/(v+c) + 1/(v-c)
(t/450) (v+c) (v-c) = (v-c) + (v+c)          (*multiplying by v-c and by v+c)
(t/450) (v²-c²) = 2v
v² - c² = (900/t) v
v² - (900/t) v - c² = 0
v = 450/t ± √((450/t)² + c²)

v = 450/t + √((450/t)² + c²)     (since the - would result in a negative value)

 

So for line 1, that's

v = 15/8 + √((15/8)² + 1²)

   = 15/8 + √(225/64 + 64/64)

   = 15/8 + √(225+64)/8
   = 15/8 + √289/8
   = 15/8 + 17/8

   = 32/8

   = 4

 

While for line 2, 450/t=450/300=3/2 so that

v = 3/2 + √((3/2)² + 2²)
= 3/2 + √(9/4 + 4)
= 3/2 + √(9/4 + 16/4)
= 3/2 + √(25/4)
= 3/2 + 5/2
= 8/2
= 4

Lines 3, 4 & 6 just need the original equation -- which needs parentheses around each denominator, but can also be simplified algebraically, if we note that (v+c)(v-c) = v²-c² for the common denominator line below:

 

t = 450/(v+c) + 450/(v-c)

  = 450( 1/(v+c) + 1/(v-c) )

  = 450( (v-c)/(v²-c²) + (v+c)/(v²-c²) )   (common denominator)

  = 450( 2v / (v²-c²) )
  = 900 v / (v² - c²)

 

and results in times of 900/(16-9)=900/7≈128.57, 900/(9-1)=900/8=112.5 & 900/0=undefined (infinite) seconds, respectively.

 

Lastly, for line 5, we need to solve for c. Taking the last equation above (where c² is isolated), we have:

 

v² - c² = 900v/t      (multiplying by v²-c² and dividing by t)

c² = v² - 900v/t      (adding c² and subtracting 900v/t)

 

so that c is the square root of this. Thus, for line 5,

 

c² = 3² - 2700 / 540 = 9 - 5 = 4

 

giving c = 2.

 

(Note that c=±√4=±2 makes sense here, since the original equation is the same with -c substituted for c and reflects an uncertainty about which direction is faster when we only know the total time in both directions.)