Aboat traveled downstream a distance of 80 mi and then came right back. if the speed of the current was 8 mph and the total trip took 5 hours and 20 minutes, find the average speed of the boat relative to the water.
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Ответ:
Velocity ot the boat downstream = b + 8
Velocity of the boat upstream = b - 8
v = d / t => t = d / v
Time downstream = 80 / [b + 8]
Time upstream = 80 / [b - 8]
Total time 80 / [b + 8] + 80 / [b - 8] = 17/3 ... [ I converted 5 h + 20 m to 1/
hours]
Now you have to solve
80 / [b + 8] + 80 / [b - 8] = 17/3
80[b - 8] + 80[b + 8] = [17/3](b+8)(b-8)
Divide by 80
b - 8 + b + 8 = [17/(80*3)] (b^2 - 64)
2b = [17/240]b^2 - 64*17/240
0.0708b^2 -2b - 4.53333 = 0
Use the quadratic formula to solve. You will obtain b = - 2.11 and b = 30.36
Only the positve result make sense. Then the answer is b = 30.36 mph.
Let's verify that that result is coherent:
Downstream velocity = 30.36 + 8 = 38.36 mi / h
Downstream time = 80mi / 38.36 mi/h = 2.085 h
Upstream velocity = 30.36 - 8 = 22.36 mi / h
Upstream time = 80 mi / 22.36 mi/h = 3.578 h
Total time = 2.085 h + 3.578 h = 5.663 h, which is 5 h + 20 min; then, the result is correct.
Ответ:
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Explanation: