Question 89938: It is the year 2074 and the electric truck is a reality. Jeff is an independent truck driver who delivers vehicles from the factory to the dealers in his very flat state. Jeff’s truck (with him and his lunch box in it) weighs 8,000 lbs. From Jeff’s house in Small Town to the automobile factory, which is 120 miles away, his truck consumes 99.86 kW-h of energy. He loads up 6 vehicles with a nominal weight each of 3000 lbs. His closest customer, dealership Rip-U-Off, is 200 miles away and Jeff’s truck consumes another 543 kW-h getting there. He delivers X trucks to Mr. Rip-U-Off and continues on to dealership U-Bought-It to deliver the rest of the vehicles. U-Bought-It is 300 miles from Rip-U-Off and Jeff’s truck consumed 436.89 kW-h to get to it. Jeff returns home with his empty truck and consumes 41.61 kW-h to get home (he was so busy that had no time to eat his lunch).
This is a challenging problem. Let’s not get frustrated trying to solve it right, but rather let’s spend time formulating it and discussing the possibilities. Of course, the easy questions are:
· How many vehicles did each dealer get? and
· How far is Jeff’s home from the U-Bought-It dealership?
The bonus question is which of the two dealers received a smuggled shipment of potatoes in the trunk of one of the vehicles and how much did the shipment weigh?
Answer by scott8148(6628)   (Show Source):
You can put this solution on YOUR website! the energy that Jeff's truck uses is proportional to the distance he drives and the weight of the load e=kdw
k is the constant of proportionality and is determined for the truck from Jeff's initial trip k=e/(dw)
k=99.86/(120*8000) ... using the equation e=kdw, you can plugin values from the legs of the trip to find the weights
Jeff's trip home at the end is with the same weight as the beginning so the distance is porportional to the energy
each dealer gets 3 cars ... Jeff's house is 50 miles from U-Bought-It ... Rip-You-Off got 100 lbs of potatoes
|
|
|
