Recent content by loldawgz

Apparently this answer of 25 mph was right...who knows but thanks guys for the help!

Yeah, I know the chain rule for derivatives. If you want to go by that and your previous relation, an equation could be s'(x(t)) * x'(t).

You plug in t for x in s(x), is what I'm assuming

So, if you look at the both the functions, x is a common variable between them, representing the number of gallons. Using this, you can find the miles per gallon at a specific time. For example, if you look at the second graph, the x(gallon) value at t = 15 hours is approximately 9. You can take...

I think I have ds/dt already, by combining the ds/dx function and the x/t function, I graphed the ds/dt function. By checking the value at t = 15, the velocity appears to be 25 mph. Is this correct?

s(t) isn't given, but I can construct a ds/dx as a function of time instead. ds/dx in this case is miles/gallon. What I'm confused with, however, is trying to create the s(t function from the ds/dx function. I tried multiplying the ds/dx by t function by the gallons given at that time, but it...

I'm pretty sure you take the derivative of the d(t) function

Homework Statement Suppose you take a car trip, traveling east along a very long highway, starting at time t 0. Let x t be the number of gallons of gasoline used during the first t hours, and let st be the distance traveled in that time. Because you’re using very cheap gasoline that’s not good...