1.1 Average Velocity

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We compute average velocity to estimate instantaneous velocity.

1 Average Velocity

Example 1 Consider an object falling from a height of 100 feet. The height of the object, in feet, after $t$ seconds is given by the function $h(t)= 100 - 16t^2$ . To find the average velocity of the object during the fall, we can use the formula $v_{ave}[a,b] = \frac {\Delta h}{\Delta t} = \frac {h(b) - h(a)}{b-a}.$

The Greek letter $\Delta$ (delta) represents “change in”, so $\Delta h$ is the change in height and $\Delta t$ is the change in time. To represent the entire 100-foot fall of the object, we will use $a = 0$ , and for $b$ , we must know how long it takes for the object to hit the ground. To determine $b$ , we set $h(t) = 0$ and solve for $t$ : $100-16t^2 = 0 \implies 16t^2 = 100 \implies t = \pm \sqrt {\frac {100}{16}} \implies t = \frac 52$ Thus the object hits the ground after $b = 2.5$ seconds, and we can compute the average velocity of the object during the fall: $v_{ave}[0, 2.5] = \frac {h(2.5) - h(0)}{2.5 - 0} = \frac {0 - 100}{2.5 - 0} = \frac {-100}{2.5} = -40 \text { ft/sec}.$ The negative sign indicates that the object is falling.

(Problem 1a) The height of a vertically free-falling object after $t$ seconds is given by $h(t) = 400 - 16t^2$ feet. Find the average velocity of the object during the fall.
The initial height of the object is $h(0) =\answer {400}$ ft
The object hits the ground after $t =\answer {5}$ sec
The average velocity during the fall is $v_{ave}[0, 5] = \answer {-80}$ ft/sec
A negative velocity indicates that the object is rising falling

(Problem 1b) The height of a vertically free-falling object, in feet after $t$ seconds is given by $h(t) = 300 - 16t^2$ feet. Find the average velocity of the object from time $t = 2$ seconds to time $t = 4$ seconds.
At times $t = 2$ and $t= 4$ , the corresponding heights are: $h(2) = \answer {236}\, \text {feet}, \; \text {and} \; h(4) = \answer {44} \,\text {feet}$ The average velocity of the object over the time interval $[2,4]$ is: $v_{ave}[2,4] = \answer {-96}\, \text {ft/sec}$ The negative sign indicates that the object is rising falling

Average velocity has a connection with the slope of a line. Recall that the slope of a line between two points, $(x_1, y_1)$ and $(x_2, y_2)$ is given by $m = \frac {\text {rise}}{\text {run}} = \frac {\Delta y}{\Delta x} = \frac {y_2 - y_1}{x_2 -x_1}.$ Thus, if we make a graph of position versus time, with time on the horizontal axis and position on the vertical axis, then the average velocity of a vertically free-falling object from time $t = a$ to time $t = b$ is the same as the slope of the line segment connecting the points $(a, h(a))$ and $(b, h(b))$ . See the figure below.

example 2 In the last example, with $s(t) = 100 - 16t^2$ , we found that $s(0) = 100$ ft and $s(5/2) = 0$ ft. If we plot the points $(0, 100)$ and $(5/2, 0)$ then the slope of the segment connecting them is the average velocity of the falling object from time $t = 0$ seconds to time $t = 5/2$ seconds, as shown in the figure below.

(problem 2) Find the slope of the line between the points $(2,236)$ and $(4,44)$ .
$m = \answer {-96}.$

2 Instantaneous Velocity

To determine the velocity of an object at a particular moment in time, i.e., the instantaneous velocity, we find the average velocity over smaller and smaller time intervals.

example 3 The height of a falling object at time $t$ seconds is given by $s(t) = 600-16t^2$ feet. Find the average velocity of the object over the time intervals $[4, 5], [4.5, 5], [4.9, 5]$ and $[5, 5.1]$ .
First we will compute the height (in feet) at time $t = 4, 4.5, 4.9, 5$ and $5.1$ seconds:

$\begin {array}{ c | c | c | c | c | c } t& 4 & 4.5 & 4.9 & 5 & 5.1 \\ \hline s(t) & 344 & 276 & 215.84 & 200 & 183.84 \end {array}$

The average velocities (in ft/sec) are: $\begin {array}{ c | c } \text {interval}& v_{ave} \\ \hline [4,5] & -144 \\ \hline [4.5,5] & -152 \\ \hline [4.9,5] & -158.4 \\ \hline [5, 5.1] & -161.6 \end {array}$

We can now estimate the instantaneous velocity of the object at time $t = 5$ seconds. Based on the average velocities over the smallest time intervals, namely, $-158.4$ ft/sec and $-161.6$ ft/sec, it seems that $-160$ ft/sec would be a reasonable estimate for the instantaneous velocity of the falling object at time $t = 5$ seconds. To improve our estimate, or have more confidence in it, we could compute the average velocity over even smaller time intervals.

(problem 3a) The height of a vertically free-falling object at time $t$ seconds is given by $s(t) = 200 - 16t^2 \; \text {feet}.$ Find the average velocity of the object over each of time intervals given below and use your results to estimate the instantaneous velocity of the object at time, $t = 3$ seconds.

$\begin {array}{ c | c } \text {interval}& v_{ave} \\ \hline [2.9, 3] & \answer {-94.4} \text {ft/sec} \\ \hline [2.99, 3] & \answer {-95.84} \text {ft/sec} \\ \hline [3, 3.1] & \answer {-97.6} \text {ft/sec} \\ \hline [3, 3.01] & \answer {-96.16} \text {ft/sec} \end {array}$

Instantaneous velocity at $t = 3$ seconds: $\answer {-96}ft/sec$ .

(problem 3b) The height of a vertical projectile at time $t$ seconds is given by $s(t) = 50t - 16t^2 \text {feet}.$ Find the average velocity of the object over each of time intervals given below and use your results to estimate the instantaneous velocity of the object at time, $t = 1$ second.

$\begin {array}{ c | c } \text {interval}& v_{ave} \\ \hline [0.9, 1] & \answer {19.6} \text {ft/sec} \\ \hline [0.99, 1] & \answer {18.16} \text {ft/sec} \\ \hline [1, 1.1] & \answer {16.4} \text {ft/sec} \\ \hline [1, 1.01] & \answer {17.84} \text {ft/sec} \end {array}$

Instantaneous velocity at $t = 1$ second: $\answer {18}\text {ft/sec}$
The positive answer indicates that the projectile was rising at time $t= 1$ second.

3 The Tangent Line

We saw that the average velocity of a falling object can be represented by the slope of a secant line. The instantaneous velocity can also be represented by the slope of a tangent line.

The function in the graph below represents the position function for a falling object. The slope of the blue secant line represents the average velocity of the object over the time interval $[a,b]$ . The slope of the red tangent line represents the instantaneous velocity of the object at time $t = b$ .

In differential calculus, we study the tangent line and its applications.

2026-01-31 23:15:57