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 = t2  Δ v+cs = vcsΔ  s(vc)=s (v+c)Δ (vc) (v+c)  900(v25)=900 (v+25)0.16666666666667 (v25) (v+25) 0.16666666667152v245104.167=0 v1,2=±45104.166666667/0.16666666667152=±520.21630116 v1=520.21630116 v2=520.21630116  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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