Pilar puts the boat in a lake that travels 400 meters at a constant speed. The time. On the return, the speed was the same for 2 minutes and then increase by 10m/min. The time boat takes in return back = 8 minutes
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 the time for the trip, t₁
[tex]t_2 = t_i - 1[/tex]
Therefore we have;
[tex]v_i\times t_i = v_i \times 2 + v_2 \times (t_2-2) \\= 400\\v_i \times 2 + (v_i + 10) \times (t_2-2) = 400\\(v_i + 10) \times t_2 - 20 = 400\\[/tex]
but
[tex]v_i = \frac {400}{t} = \frac{400)}{t_2+ 1}[/tex]
Which gives;
[tex]\frac{(400) }{(t₂ + 1} + 10) \times t_2 - 20 = 400\\10\times ( t_2²+ 36·t_2-2)/(t_2+1) = 40010 \times t_2^²+ 10 \times t_2-420 = 0[/tex]
[tex]t_2^2+ t_2-42 = 0[/tex]
[tex](t_2 - 7)(t_2 + 6) = 0[/tex]
[tex]t_2 = 7[/tex] minutes or [tex]-6[/tex] minutes
Given that t₂ is a natural number, we have,
t₂ = 7 minutes
Whereby, [tex]t_2 = t_i - 1[/tex], we have;
[tex]7 = t_i - 1[/tex]
[tex]t_{i} = 1 + 7 = 8[/tex] Minutes
The trip normally takes 8 minutes
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