Asked by Delilah
Two boat landings are 8.0 km apart on the same bank of a stream that flows at 2.8 km/h. A motorboat makes the round trip between the two landings in 50 minutes. What is the speed of the boat relative to the water?
I think it is something along the lines of
4/t+2.8 + 4/t-2.8 =5/6, but when I solve for x, I get the incorrect answer. Could you help me out?
I think it is something along the lines of
4/t+2.8 + 4/t-2.8 =5/6, but when I solve for x, I get the incorrect answer. Could you help me out?
Answers
Answered by
Henry
T = 50min * 1h/60min = 5/6 h.
T = t1 + t2 = 5/6 h.
d/(Vb-Vs) + d/(Vb+Vs) = T
8/(Vb-2.8) + 8/(Vb+2.8) = 5/6
LCD = (Vb-2.8)(Vb+2.8).
(8(Vb+2.8)+8(Vb-2.8))/(Vb-2.8)(Vb+2.8)=
5/6.
(8Vb+22.4+8Vb-22.4)/(Vb-2.8)(Vb+2.8)=5/6
16Vb/(Vb-2.8)(Vb+2.8) = 5/6
Cross multiply:
96Vb = 5(Vb-2.8)(Vb+2.8)
96Vb = 5(Vb^2-2.8^2)
96Vb = 5Vb^2-39.2
-5Vb^2 + 96Vb + 39.2 = 0
Use Quadratic formula and get:
Vb = 19.6 km/h = Speed of the boat.
T = t1 + t2 = 5/6 h.
d/(Vb-Vs) + d/(Vb+Vs) = T
8/(Vb-2.8) + 8/(Vb+2.8) = 5/6
LCD = (Vb-2.8)(Vb+2.8).
(8(Vb+2.8)+8(Vb-2.8))/(Vb-2.8)(Vb+2.8)=
5/6.
(8Vb+22.4+8Vb-22.4)/(Vb-2.8)(Vb+2.8)=5/6
16Vb/(Vb-2.8)(Vb+2.8) = 5/6
Cross multiply:
96Vb = 5(Vb-2.8)(Vb+2.8)
96Vb = 5(Vb^2-2.8^2)
96Vb = 5Vb^2-39.2
-5Vb^2 + 96Vb + 39.2 = 0
Use Quadratic formula and get:
Vb = 19.6 km/h = Speed of the boat.
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