Instantaneous Velocity at t=4: Solving Displacement Equation s= t^2- 5t + 15

[Amber430]

Q: Displacement in a straight line is s= t^2- 5t + 15. What's the instantaneous velocity when t= 4?


Equation: lim [f(a+h)-f(a)]/h as h approaches 0

I tried this problem and what I came up with for an idea of how to solve this makes absolutely no sense.

 

[Russell Berty]

Have you covered rules for derivatives yet?

 

[Cyosis]

Even if you think your method of solution makes absolutely no sense you should still tell us what you did. This way we can see what the real problem is.

Do you understand what the formula you were given represents? Normally when you're given a distance function and you want to calculate the average velocity between two time intervals you will calculate s2-s1 divided by the time it took. The smaller you take this "width" of time the closer you get to the instantaneous velocity. Does it make sense now?

 

[Mark44]

To expand just a bit on what Cyosis wrote, and looking at your displacement as a function of t, you have s(t) = t^2- 5t + 15.

What you need to do is to calculate
\lim_{h \rightarrow 0} \frac{s(4 + h) - s(4)}{h}
That will be your instantaneous velocity at t = 4.