Instantaneous velocity

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We use limits to compute instantaneous velocity.

Video Lecture

Text (and Concept Questions)

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 $s$ .

1 : When will the ball hit the ground?

To determine when the ball hits the ground we need to know when the distance to the ground is 0. In other words, we want 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 a time when $t>0$ .

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

2 : 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.

3 : Which points correspond to the height of the ball at time $t=0$ , $t=1$ and $t=2$ seconds?

(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.

4 : 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.

5 : 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.

6 : 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 $v_{\text {av}} =\answer [given]{8-16t}\hspace {0.06in}\unit {ft}/\unit {s}.$

7 : 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 is 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 $\answer [given]{2}$ . $\begin{align*} v (2)= \lim_{t\to2}v_{\text{av}} =\lim_{t\to2} \frac{s(t)-s(2)}{t-2} &= \lim_{t\to2}(8-16t)\\ &= \answer[given]{-24}\unit{ft}/\unit{s}. \end{align*}$