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

\[ t=\answer [given]{2.5} \unit {seconds.} \]

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 $0$. As $t$ approaches $2$, we are computing the average velocity on smaller and smaller time intervals, and the limit of these average velocities is called the instantaneous velocity at $t=2$. Limits will allow us to compute instantaneous velocity.

As before, suppose $s(t)$ denotes the position, at the time $t$, of an object moving along a line. Let $v_{\text {av}}(t)$ be the average velocity on the interval $[a, t]$ if $t>a$ and on interval $[t, a]$ if $t<a$.

The instantaneous velocity at the time $t=a$ is given by

\[ v(a) = \lim _{t\to a} v_{\text {av}}(t) = \lim _{t\to a}\dfrac {s(t)-s(a)}{t-a}.\]

The instantaneous velocity is given as a limit, $v(a) = \lim _{t\to a}\dfrac {s(t)-s(a)}{t-a}$. What is the form of this limit? The numerator $s(t)-s(a)$ approaches $0$ as $t\to a$ (since the displacement function $s(t)$ is continuous) and the denominator $t-a$ also approaches $0$. This limit has form $\relax \boldsymbol {\tfrac {0}{0}}$, an indeterminate form.

Let’s use the same setting as before to work with this definition.

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

\[ v_{\text {av}}(t) = \frac {s(t)-s(2)}{t-2}= 8-16t \hspace {0.06in}\unit {ft}/\unit {s}. \]

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\to 2}v_{\text {av}}(t)\\ &=\lim _{t\to 2} \frac {s(t)-s(2)}{t-2} \\ &= \lim _{t\to 2}(8-16t)\\ &= \answer [given]{-24}\unit {ft}/\unit {s}. \end{align*}