1. Calc Word problem

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

2. The average velocity is the gradient of the line that joins $\displaystyle (0,2)$ and $\displaystyle (1,2)$.

The instantaneous velocity is the gradient of a line that joins any point to another point extremely close to it. So for any point $\displaystyle (x, f(x))$, you can say that a point close to it is $\displaystyle (x + h, f(x+h))$, and the gradient would be $\displaystyle \frac{f(x+h) - f(x)}{x+h-x} = \frac{f(x+h) - f(x)}{h}$.

So in your question, to evaluate the instantaneous velocity at $\displaystyle t = 1$, you know that on the curve, when $\displaystyle t = 1, f(t) = f(1) = 62$. A point close to this would be $t = 1 + h$, so $\displaystyle f(t) = f(1 + h) = 2(1 + h)^2 + 60(1 + h)$.