SOLUTION: Please help me.
A truck travels 450 km to a delivery point, unloads, and, now empty, returns to the starting point at a speed 8.00 km/h greater than on the outward trip. What was
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A truck travels 450 km to a delivery point, unloads, and, now empty, returns to the starting point at a speed 8.00 km/h greater than on the outward trip. What was
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Question 1057917: Please help me.
A truck travels 450 km to a delivery point, unloads, and, now empty, returns to the starting point at a speed 8.00 km/h greater than on the outward trip. What was the speed of the outward trip if the total round-trip driving time was 14.4 h? Keep three significant digits.
Upper R 1 =
km/h
You can put this solution on YOUR website! Let s be the speed of the truck going toward the delivery point. Then the return trip would be s+8. So:
450/s + 450/s+8=14.4
450(s+8)+450s=14.4sē+115.2s
14.4sē-784.8s-3600=0
144sē-7848s-36000=0
Using the quadratic formula, we get two roots as a solution:-4.25495992697 and 58.754959927. Discarding the negative result, we get the outward trip speed as 58.755 kph. ☺☺☺☺
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