# find average velocity and instantaneous velocity [closed]

The displacement (in meters) of an object moving in a straight line is given by:

$$f(t)= 1 + 2t + \frac{1}{4}t^2$$

where $t$ is measured in seconds.

Find the average velocity over the following time periods:

(i) $[1, 1.2]$

(ii) $[1, 1.1]$

(iii) $[1, 1.01]$

(iv) $[1, 1.001]$

Also use the information above to estimate the instantaneous velocity when $t = 1$ second.

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## closed as off-topic by 900 sit-ups a day, Gina, hardmath, le gâteau au fromage, anortonJul 19 at 0:32

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I will assume that this is homework.

Average velocity is determined by dividing the distance travelled by the time it takes to travel that far. Therefore, for the interval $[ 1, 1.2 ]$ we have $$\dfrac{\textrm{distance}}{\textrm{time}} = \dfrac{f(1.2) - f(1)}{1.2 - 1} = \dfrac{3.76 - 3.25}{0.2} = 2.55 ~m/sec.$$

The other parts are quite similar. To estimate the instantaneous velocity at 1, we estimate what the values you receive from parts i-iv seem to be getting closer to.

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Average velocity of $f(t)$ between $t=a$ and $t=b$ is simply the slope of the line that travels through these points. Thus

$$\frac{\Delta d}{\Delta t}=\frac{f(1.2)-f(1)}{1.2-1}$$

Instantaneous velocity at a point $t = p$ is the same as average velocity where the two points are both very close to $p$, i.e.

$$\lim_{\Delta t \to 0}\frac{\Delta d}{\Delta t}= \lim_{\Delta t \to 0}\frac{f(1 + \Delta t)-f(1)}{\Delta t}$$

So you may thus choose $\Delta t \approx 0$ to get an approximation, or, if you know how, differentiate $f$ at $1$.

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