A speedboat

A speedboat goes 900 km against a current of 25 km/hr. Its trip took 10 minutes longer than it would have taken to travel with the current. What is the speedboat's speed in still water?

Correct answer:

v =  520.2163 km/h

Step-by-step explanation:

s=900 km c=25 km/h Δ=10 min h=10:60  h=0.16667 h  t1=v+cs t2=vcs  t1=t2Δ v+cs=vcsΔ  s(vc)=s(v+c)Δ(vc)(v+c)  900(v25)=900 (v+25)0.16666666666667 (v25) (v+25) 0.166667v245104.167=0  a=0.166667;b=0;c=45104.167 D=b24ac=0240.166667(45104.167)=30069.504583333 D>0  v1,2=2ab±D=0.333334±30069.5 v1,2=±520.2157809516 v1=520.2157809516 v2=520.2157809516   Factored form of the equation:  0.166667(v520.2157809516)(v+520.2157809516)=0  v>0  v=v1=520.2163=520.2163 km/h

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Showing 1 comment:
Campo
t1+t2=T
900/(v-25)+900/(v+25)==2(900/v)+10/60
Ans. 190.09kph
note:
1. the whole trip is 2 times the time plus 10mins due to counter current. (time= 2T + 10/60). time unit is in hours.
2. (900/x) is the velocity in still waters. since Distance = V*T, thus, T=D/v.
hope it can helps.





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