Calc Word problem

**agent2421** · February 12th 2011, 06:11 PM

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

**Prove It** · February 12th 2011, 06:17 PM

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 $\displaystyle f(t) = f(1 + h) = 2(1 + h)^2 + 60(1 + h)$ .

Simplify then put into your equation for the gradient.

**agent2421** · February 12th 2011, 07:01 PM

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

**Prove It** · February 12th 2011, 07:05 PM

Gradient means slope...

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