Two trains, Train A and Train B, weigh a total of 150 tons. Train A is heavier than train B. The difference of their weights is
2 answers:
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The river flows at 2 miles per hour. This is the speed of the current, or the speed at which the water itself is moving.
The mass of the particles moving in river is proportional to the sixth power of 2, or
miles per hour. So, if x is the mass of the particle, then the speed is modeled by an inverse relationship 
, for some constant k.
For particles that are 15 times the usual mass, the speed of the river itself must increase for the particles to travel at the same speed as before. The mass of the particles is still proportional to the sixth power of the speed of the river, but the speed of the river can no longer be 2 miles per hour. It is some speed y, which we are solving for.
So, the key here is proportional. We can set up two proportions and cross multiply.
Cancelling our x's (because we don't need to solve for them), we have
To solve for y, we must take the sixth root of both sides, or raise both sides to the one-sixth power.
![\sqrt[6]{y^6} = \sqrt[6]{960}](https://tex.z-dn.net/?f=%20%5Csqrt%5B6%5D%7By%5E6%7D%20%3D%20%20%5Csqrt%5B6%5D%7B960%7D%20)
Using a calculator, we see that![\sqrt[6]{960} = 3.14 \ miles \ per \ hour](https://tex.z-dn.net/?f=%5Csqrt%5B6%5D%7B960%7D%20%3D%203.14%20%5C%20miles%20%5C%20per%20%5C%20hour)
.
The mass of the particles moving in river is proportional to the sixth power of 2, or
For particles that are 15 times the usual mass, the speed of the river itself must increase for the particles to travel at the same speed as before. The mass of the particles is still proportional to the sixth power of the speed of the river, but the speed of the river can no longer be 2 miles per hour. It is some speed y, which we are solving for.
So, the key here is proportional. We can set up two proportions and cross multiply.
Cancelling our x's (because we don't need to solve for them), we have
To solve for y, we must take the sixth root of both sides, or raise both sides to the one-sixth power.
Using a calculator, we see that