Answer:
The trip normally takes 8 minutes
Step-by-step explanation:
The given information states that the away distance the boat traveled = 400 m
The time traveled at the same initial speed , v₁, by the boat on the way back = 2 minutes
The increase in speed of the boat by Pilar = 10 m/min
The new speed, v₂ = v₁ + 10
The time for the return trip, t₂ = 60 seconds (1 minute) faster than time for the trip, t₁
t₂ = t₁ - 1
Therefore we have;
v₁ × t₁ = v₁×2 + v₂×(t₂-2) = 400
v₁×2 + (v₁ + 10)×(t₂-2) = 400
(v₁ + 10)×t₂ - 20 = 400
But v₁ = 400/t₁ = 400/(t₂ + 1)
Which gives;
(400/(t₂ + 1) + 10)×t₂ - 20 = 400
10×(t₂²+ 36·t₂-2)/(t₂+1) = 400
10·t₂²+ 10·t₂-420 = 0
t₂²+ t₂-42 = 0
(t₂ - 7)(t₂ + 6) = 0
t₂ = 7 minutes or -6 minutes
Given that t₂ is a natural number, we have, t₂ = 7 minutes
Whereby, t₂ = t₁ - 1, we have;
7 = t₁ - 1
t₁ = 1 + 7 = 8 Minutes
The trip normally takes 8 minutes
Answer:
actually, it is 7
Step-by-step explanation:
the previous guy's explanation was all correct, but the question is asking for the return trip, which is one minute less, so the answer is 7. Hope this helped, also, i go to rsm as well and i got this one correct
