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 an 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 $s(t) = 40t-16t^2 \qquad \text {(feet above the ground)} .$

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 has solutions $t=0$ seconds and $t=\answer [given]{2.5}$ seconds. Since the ball hits the ground it’s thrown, we know that the ball hits the ground at $t=\answer [given]{2.5}$ seconds.

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

To find the height of the ball after $2$ seconds we simply need to plug $2$ into the equation for $s(t)$ to find $s(2) = 40\cdot \answer [given]{2} - 16\cdot \answer [given]{2}^2 = \answer [given]{16}\unit {ft.}$

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

Which points correspond to the height of the ball at times $0$ , $1$ and $2$ ?

The point that corresponds to $s(0)$ , the position (height) of the ball at $t=0$ , is ABCD .

The point that corresponds to $s(1)$ , the position (height) of the ball at $t=1$ , is ABCD .

The point that corresponds to $s(2)$ , the position (height) of the ball at $t=2$ , is ABCD .

Next let’s consider the average velocity of the ball. What is the average velocity of the ball on the 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 interval $[a,b]$ is given by $v_{\text {av}} = \frac {s(b) - s\left (\answer [given]{a}\right )} {\answer [given]{b} - \answer [given]{a}}.$ Plugging in $a=1$ and $b=2$ we find that $v_{\text {av}} = \frac {s(2)-s(1)}{2-1} = \frac {\answer [given]{16} - \answer [given]{24}}{1} = \answer [given]{-8}\unit {ft}/\unit {s}.$

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

We use the same formula we used to find the average velocity on the interval $[1,2]$ to find the average velocity on the interval $[t,2]$ for $0<t<2$ .

$\begin{align*} v_{\text{av}} &= \frac{s(2)-s\left(\answer[given]{t}\right)} {\answer[given]{2}-\answer[given]{t}}\\ &= \frac{16 - (40t-16t^2)}{2-t}\\ &= \frac{8(2t^2-5t+2)}{2-t}\\ &= \frac{8(2t-1)(t-2)}{-(t-2)}\\ &= -8(2t-1)\\ &= (8-16t)\unit{ft}/\unit{s} \end{align*}$

for $0<t<2$ .

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

To calculate the average velocity on the interval $[2,t]$ for $2<t<2.5$ we will use our average velocity formula one more time to find $v_{\text {av}} = \frac {s(t)-s(2)}{t-2} = \frac {s(2)-s(t)}{2-t}.$ However, this is exactly the same expression we got when calculating the average velocity on the interval $[t,2]$ for $0<t<2$ . So the average velocity on the interval $[2,t]$ for $2<t<2.5$ is given by $v_{\text {av}} =\answer [given]{(8-16t)}\unit {ft}/\unit {s}.$

In our previous example, we computed average velocity on several different intervals. If we let the size of the interval go to zero, we get instantaneous velocity. Limits will allow us to compute instantaneous velocity. Let’s use the same setting as before.

The height of a ball above the ground between $0$ and $2.5$ seconds is given by $s(t) = 40t - 16t^2.$ Find 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 $v_{\text {av}} = \frac {s(t)-s(2)}{t-2}= \answer [given]{(8-16t)}\unit {ft}/\unit {s}.$ All we need to do to find the instantaneous velocity at $t=2$ is take the limit as $t$ goes to $\answer [given]{2}$ of the expression above. Doing so we find $\begin{align*} v = \lim_{t\to2}v_{\text{av}} =\lim_{t\to2} \frac{s(t)-s(2)}{t-2} &= \lim_{t\to2}(8-16t)\\ &= \answer[given]{-24}\unit{ft}/\unit{s}. \end{align*}$