# Confused on next step in finding instantaneous velocity

## Homework Statement

In a time of t seconds, a particle moves a distance of s meters from its starting point, where s=3t$$^2$$+13.

Find the average velocity between t=1 and t=1+h if
(i) h=0.1 (ii) h=0.01 (iii) h=0.001

Use the obtained answers to estimate the instantaneous velocity of the particle at time t=1.

The average velocity between t=1 and t=1+h
(i) when h=0.1 is ____ m/s;
(ii) when h=0.01 is ____ m/s;
(iii) when h=0.001 is ____ m/s.

The instantaneous velocity appears to be ______ m/s

## The Attempt at a Solution

h=0.1 3(0.1)$$^2$$+13=13.03
h=0.01 3(0.01)$$^2$$+13=13.0003
h=0.001 3(0.001)$$^2$$+13=13

What am I supposed to do next in this problem?

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Dick
Homework Helper
You'd better back up and rethink the first part. The average speed between t=1 and t=1+h is s(1+h)-s(1) divided by the elapsed time (h). It's just distance divided by time.

You'd better back up and rethink the first part. The average speed between t=1 and t=1+h is s(1+h)-s(1) divided by the elapsed time (h). It's just distance divided by time.
If I do s(1+0.1)-s(1)/0.1 that gives me s(0.1)/0.1

I don't get it. How does that work?

Dick
Homework Helper
If I do s(1+0.1)-s(1)/0.1 that gives me s(0.1)/0.1

I don't get it. How does that work?
Nooo. s(1+0.01)-s(1) is NOT s(0.01). s(1+0.01)-s(1)=(3*(1.01)^2+13)-(3*1^2+13). That's not the same as (3*(0.01)^2+13). Check the two expressions using your calculator!

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I got s(0.1)=0.63, s(0.01)=0.0603, and s(0.001) was too small a number it appeared 0. I don't get how this gets the instantaneous velocity. Do you multiply by 10 and 100 and 1000.. etc..?

Dick