This could be made simpler in practice. We need to find the instantaneous rate of additional energy/weight gained by taking the first derivative of your equation.

But, what if the service is really really slow. Like I woulda gotten it faster if I woulda jumped the counter and made it myself. Cuz this one time it took them over a half an hour to make an order and if I wasn't in the middle of a food desert I woulda left and gotten food somewhere else.

## Replies

5,834Member Member Posts:5,834MemberThis could be made simpler in practice. We need to find the instantaneous rate of additional energy/weight gained by taking the first derivative of your equation.

f(E)= ((mc^2)^2+((pc)^2)/(E^2))

f ' (E) = (2 ( (mc^2+pcE^2) - (E(mc^2)^2 + ((pc^2) / 2) / E^4)

So for any given food and momentum we should be able to find the precise rate of weight gain.

443Member Member Posts:443Member443Member Member Posts:443MemberOh my, that IS a nice profile pic...

919Member Member Posts:919Member12,620Member Member Posts:12,620Member1,150Member Member Posts:1,150Member