Question 68171This question is from textbook College Algebra
: A speedboat takes 1 hour longer to go 24 miles up a river than to return. If the boat cruises at 10 miles per hour in still water, what is the rate of the current?
If R*T=D, and R is speed + or - current, how can you find the rate without the time, and how can you find the time without the rate?
This question is from textbook College Algebra
Answer by ptaylor(2198)   (Show Source):
You can put this solution on YOUR website!
If R*T=D, and R is speed + or - current, how can you find the rate without the time, and how can you find the time without the rate?
THE KEY WAS WHEN WE WERE TOLD THAT IT TAKES ONE HOUR LONGER TO GO UPSTREAM THAN DOWNSTREAM BECAUSE T=D/R AND WE ARE TOLD THAT:
D/R(Upstream) minus 1 equals D/R (DOWNSTREAM)
Let x=rate of current
distance(d)=rate(r) times time(t) or d=rt or t=d/r
distance upstream=distance downstream=24 mi
rate upstream=10-x
rate downstream=10+x
time upstream=(distance upstream)/(rate upstream)=24/(10-x)
time downstream=(distance downstream)/(rate downstream)=24/(10+x)
Now, we are told that time upstream minus one hour equals time downstream. So our equation to solve is:
24/(10-x)-1=24/(10+x) Multiply both sides by (10-x)(10+x)
24(10+x)(10-x)/(10-x)-1(10-x)(10+x)=24(10-x)/(10+x)/(10+x) clear fractions
24(10+x)-(10-x)(10+x)=24(10-x) clear parens
240+24x-100-10x+10x+x^2=240-24x subtract 240 from and add 24x to both sides
240-240+24x+24x-100+x^2=240-240-24x+24x collect like terms
x^2+48x-100=0 ----------------quadratic equation in standard form
This equation can be factored:
(x+50)(x-2)=0
x=-50 mph----------------discount the negative value for speed
and
x=2 mph----------------------speed of the current
CK
10-x=10-2= 8 mph -----------------------speed upstream
10+x=10+2=12 mph--------------------------speed downstream
time upstream=24/8=3 hours
time downstream=24/12=2 hours
so 3-2=1----takes 1 hour longer upstream
Hope this helps -----ptaylor
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