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​ = v−cs​  t1​ = t2​ − Δ v+cs​ = v−cs​−Δ  s(v−c)=s⋅ (v+c)−Δ⋅ (v−c)⋅ (v+c)  900(v−25)=900⋅ (v+25)−0.16666666666667⋅ (v−25)⋅ (v+25) 0.16666666667152v2−45104.167=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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