The accompanying graph shows the total distance s traveled...

The accompanying graph shows the total distance s traveled by a bicyclist after $t$ hours. GRAPH CANT COPY a. Estimate the bicyclist's average speed over the time intervals $[0,1],[1,2.5],$ and [2.5,3.5] b. Estimate the bicyclist's instantaneous speed at the times $t=\frac{1}{2}$ $t=2,$ and $t=3$ c. Estimate the bicyclist's maximum speed and the specific time at which it occurs.

The accompanying graph shows the total distance s traveled by a bicyclist after $t$ hours.
GRAPH CANT COPY
a. Estimate the bicyclist's average speed over the time intervals $[0,1],[1,2.5],$ and [2.5,3.5]
b. Estimate the bicyclist's instantaneous speed at the times $t=\frac{1}{2}$ $t=2,$ and $t=3$
c. Estimate the bicyclist's maximum speed and the specific time
at which it occurs.

Key Concepts

Average Rate of Change

This concept refers to the overall change in a function’s value divided by the change in the input over a specific interval. In the context of motion, when computing average speed, it involves dividing the change in distance by the change in time over a given time period, providing a measure of the overall speed during that interval.

Instantaneous Rate of Change

This idea describes the value of the derivative of a function at a specific point, which in the context of motion corresponds to the speed at a particular moment. It is interpreted as the slope of the tangent line to the graph of the distance function at that point, representing how quickly the distance is changing at that exact time.

Tangent Line and Derivative

A tangent line to a curve at a certain point touches the curve without crossing it locally, and the slope of this tangent line is the derivative at that point. This concept is fundamental in determining instantaneous rates of change, which are essential for estimating speeds at specific moments from a distance versus time graph.

Optimization and Maximum Values

This concept involves analyzing a function to determine its highest or lowest point within a given domain. In the setting of a motion graph, it translates into finding the maximum instantaneous speed by examining where the derivative achieves its greatest value, illustrating where the object's speed is at its peak.

Transcript

00:01 Asked to answer a question about the total distance traveled by a bicyclist using rates of change and tangents occurs.

00:10 So we're given the graph in exercise 21 that shows the total distance s traveled by the bicyclist after t hours.

00:19 And in part a, we are asked to estimate the bicyclist's average speed over different time intervals, including from 0 to 1, 1 to 2 .5, and 2 .5 to 3 .5.

00:34 Now to estimate average speeds over given intervals, you first need to estimate where points are located.

00:43 So we appear to have the points on the graph, 0 ,0, 1, 15, 2 .5, 20, and 3 .530 about.

01:11 Find the average speeds.

01:13 We use the equation that the average speed is the final speed, s2, minus initial speed, s2, over the 5 .30.

01:24 Final time t2 minus the initial time t1.

01:30 Therefore, for the time interval 01, the average speed is equal to 15 minus 0 over 1 minus 0, which is simply 15, and the units are in miles per hour.

02:03 Likewise, for the interval from 1 to 2 .5, the average speed of bicyclist is 20 minus 15 over 2 .5 minus 1, which is 5 over 1 .5, which is about 3 .3, and the units are miles per hour.

02:43 Finally, for the interval 2 .5 to 3 .5, the average speed of the bicyclist is equal to 30 minus 20.

03:00 Over 2 .5 minus 1 .5, sorry, 3 .5 minus 2 .5.

03:12 This is the same as 10 over 1, which of course is just 10 miles per hour.

03:23 Then in part b, we're asked to estimate the bicyclist's instantaneous speed at different times, particularly at times t equals 1⁄2 and t equals 3.

03:42 To estimate instantaneous speed, we should draw tangent lines at course, points, and then we should estimate the slope of the tangent line.

04:18 In particular, we draw some tangent lines in the graph...

Please give Ace some feedback