It took a crew 80 minutes to row 3km upstream and back again. If the rate of flow of the stream was 3km/h, what was the rowing rate of the crew?

1 Answer
Jun 10, 2016

Answer:

#-9/4+(5sqrt(7))/4color(white)(..) (Km)/h# as an exact value

#1.057 color(white)(..)(Km)/h" "# ( to 3 decimal places) as an approximate value

Explanation:

It is important to keep the units all the same.

As unit time for velocities is in hours:
Total time = 80 minutes #->80/60 hours#

Given that distance 1 way is 3Km

Let rowing velocity be #r#

Let time to row against current be #t_a#
Let time to row with current be #t_w#

Thus #t_w+t_a=80/60#

Known: distance is velocity x time
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Thus
For 'with current' #" "3Km = (r+3)t_w" "->" " t_w=3/(r+3)#
For against current#" "3Km=(r-3)t_a" "->" "t_a=3/(r-3)#

But #t_w+t_a=80/60#

#=>3/(r-3)+3/(r+3)=80/60#
'........................................................................
Consider that #a^2-b^2=(a+b)(a-b)#
'.........................................................................

#=>(3(r+3)+3(r-3))/((r-3)(r+3) )" "->" "(6r)/(r^2-9)=80/60#

#=>(360r)/80=r^2-9#

#=>r^2-(360r)/80-9=0" note that "(360/80-=9/2)#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Compare to #y=ax^2+bx+c" where "x=(-b+-sqrt(b^2-4ac))/(2a)#

#r=(-9/2+-sqrt(81/4-4(1)(-9)))/(2(1))#

#r=(-9/2+-sqrt(81/4+36))/(2)#

#r=(-9/2+-sqrt(225/4))/(2)#

#r=(-9/2+-sqrt(5^2xx7)/2)/2#

#r=-9/4+-(5sqrt(7))/4#

#=>r~~1.057" and "-3.077 " "(Km)/h#

The negative solution is not logical so

Rowing speed is:

#-9/4+(5sqrt(7))/4color(white)(..) (Km)/h# as an exact value

#1.057 color(white)(..)(Km)/h" "# ( to 3 decimal places) as an approximate value