
Calc Word problem
Can someone please help me out.
Suppose a car is moving along straight road and distanct traveled t hours after starting from rest is given by f (t) = 2t^2+60t. Find the avg velocity of the car over the time intervals (0,2) and (1,2). Find the instanteaneous velocity at t = 1 and t = 2.
Can someone explain hte process of how to solve this? Using the method equations of
f (x+h)  f(x) / h
Thanks :)

The average velocity is the gradient of the line that joins $\displaystyle \displaystyle (0,2)$ and $\displaystyle \displaystyle (1,2)$.
The instantaneous velocity is the gradient of a line that joins any point to another point extremely close to it. So for any point $\displaystyle \displaystyle (x, f(x))$, you can say that a point close to it is $\displaystyle \displaystyle (x + h, f(x+h))$, and the gradient would be $\displaystyle \displaystyle \frac{f(x+h)  f(x)}{x+hx} = \frac{f(x+h)  f(x)}{h}$.
So in your question, to evaluate the instantaneous velocity at $\displaystyle \displaystyle t = 1$, you know that on the curve, when $\displaystyle \displaystyle t = 1, f(t) = f(1) = 62$. A point close to this would be $\displaystyle t = 1 + h$, so $\displaystyle \displaystyle f(t) = f(1 + h) = 2(1 + h)^2 + 60(1 + h)$.
Simplify then put into your equation for the gradient.

Sorry, but what does gradient mean? I'm still not sure how to solve the problem for the average one at least.
