Instantaneous velocity

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We use limits to compute instantaneous velocity.

When we compute average velocity, we look at $\frac {\text {change in position}}{\text {change in time}}.$ To obtain the (instantaneous) velocity, we want the change in time to “go to” zero. By this point we should know that “go to” is a buzz-word for a limit. The change in time is often given as the length of a time interval, and this length goes to zero.

The average velocity on the (time) interval $[a,b]$ is given by $v_{\text {av}} = \frac {\text {change in position}}{\text {change in time}} = \frac {s(b)-s(a)}{b-a}.$ Here $s(t)$ denotes the position, at the time $t$ , of an object moving along a line.

Let’s put all of this together by working an example.

A young mathematician throws a ball straight into the air with a velocity of 40ft/sec. Its height (in feet) after $t$ seconds is given by $s(t) = 40t-16t^2 \qquad \text {(feet above the ground)} .$ Here is the graph of $s$ .

When will the ball hit the ground?

To determine when the ball hits the ground we need to solve the equation $s(t)=0$ for t. That is, $\begin{align*} 40t-16t^2 &= 0\\ t(40-16t) &= 0. \end{align*}$ This equation has two solutions, one of them is $t=0$ seconds.

Since the ball hits the ground some time after it’s thrown, we conclude that the ball hits the ground at time when $t>0$ .

The ball will hit ground at the time $t=\answer [given]{2.5}seconds.$

What is the height of the ball after $2$ seconds?

In order to find the height of the ball after $2$ seconds, we simply need to plug $2$ into the equation for $s(t)$ .

The height of the ball after $2$ seconds is $s(2) = 40\cdot \answer [given]{2} - 16\cdot \answer [given]{2}^2 = \answer [given]{16}\unit {ft.}$

Consider the following points lying along the $s$ axis.

Which points correspond to the height of the ball at time $t=0$ , $t=1$ and $t=2$ seconds? Make correct choices!

(a): The point that corresponds to $s(0)$ , the position (height) of the ball at $t=0$ , is
A B C D
(b): The point that corresponds to $s(1)$ , the position (height) of the ball at $t=1$ , is
A B C D
(c): The point that corresponds to $s(2)$ , the position (height) of the ball at $t=2$ , is
A B C D

Next let’s consider the average velocity of the ball.

What is the average velocity of the ball on the time interval $[1,2]$ ?

In order to find the average velocity of the ball on the interval $[1,2]$ we recall that the average velocity on the time interval $[a,b]$ is given by $v_{\text {av}} = \frac {s(b) - s\left ({a}\right )} {b-a}.$ Now we just plug in $a=1$ and $b=2$ .

$v_{\text {av}} = \frac {s(2)-s(1)}{2-1} = \frac {\answer [given]{16} - \answer [given]{24}}{1} = \answer [given]{-8}\unit {ft}/\unit {s}.$

Check the figure below.

What is the average velocity of the ball on the time interval $[t,2]$ , for $0<t<2$ ?

We use the formula for average velocity . $\begin{align*} v_{\text{av}} &= \frac{s(2)-s\left({t}\right)} {{2}-{t}}\\ &= \frac{16 - (40t-16t^2)}{2-t}\\ &= \frac{8(2t^2-5t+2)}{2-t}\\ &= \frac{8(2t-1)\cancel{(t-2)}}{-\cancel{(t-2)}}\\ &= -8(2t-1)\\ &= 8-16t \hspace{0.06in}\unit{ft}/\unit{s}, \end{align*}$ for $0<t<2$ . Check the figure below.

What is the average velocity of the ball on the interval $[2,t]$ , for $2<t<2.5$ ?

The average velocity on the interval $[2,t]$ , for $2<t<2.5$ is $v_{\text {av}} = \frac {s(t)-s(2)}{t-2} = \frac {s(2)-s(t)}{2-t}.$ Notice that this is exactly the same expression we got when calculating the average velocity on the interval $[t,2]$ for $0<t<2$ .

The average velocity on the interval $[2,t]$ , for $2<t<2.5$ , is given by $v_{\text {av}} =\answer [given]{8-16t}\hspace {0.06in}\unit {ft}/\unit {s}.$

Using the results in Questions 5 and 6, compute the average velocity on the interval

(a): $[2, 2.1]$ $v_{\text {av}} = 8-16(\answer [given]{2.1})=-25.6 \hspace {0.06in}\unit {ft}/\unit {s}$
(b): $[1.9,2]$ $v_{\text {av}} = 8-16(\answer [given]{1.9})=-22.4 \hspace {0.06in}\unit {ft}/\unit {s}$

In our previous example, we computed average velocity on several different intervals.

For example, the average velocity on the time interval $[2,t]$ is $v_{\text {av}}=8-16t$ .

Note that the size or the length of that time interval is $t-2$ .

If we let $t\to 2$ , the size of the interval will go to $0$ .

So, as $t$ approaches $2$ , we are computing the average velocity on smaller and

smaller time intervals, and the limit of these average velocities should be called

the instantaneous velocity at $t=2$ .

Limits will allow us to compute instantaneous velocity.

Let’s use the same setting as before.

The height of a ball above the ground during the time interval $[0,2.5]$ , with $t$ is in seconds and $s(t)$ in feet, is given by $s(t) = 40t - 16t^2.$ Find $v(2)$ , the instantaneous velocity of the ball $2$ seconds after it is thrown.

From the previous example, we know that the average velocity of the ball on the interval $[t,2]$ , for $0<t<2$ , and the average velocity on the interval $[2,t]$ , for $2<t<2.5$ , are both given by $v_{\text {av}} = \frac {s(t)-s(2)}{t-2}= 8-16t \hspace {0.06in}\unit {ft}/\unit {s}.$ In order to find the instantaneous velocity $v(2)$ , we take the limit as $t$ goes to $\answer [given]{2}$ . $\begin{align*} v (2)= \lim_{t\to2}v_{\text{av}} =\lim_{t\to2} \frac{s(t)-s(2)}{t-2} &= \lim_{t\to2}(8-16t)\\ &= \answer[given]{-24}\unit{ft}/\unit{s}. \end{align*}$