Compute your averagc velocity in the following two cascs:...

Compute your averagc velocity in the following two cascs: (a) You walk $73.2 \mathrm{~m}$ at a speed of $1.22 \mathrm{~m} / \mathrm{s}$ and then run $73.2 \mathrm{~m}$ at a speed of $3.05 \mathrm{~m} / \mathrm{s}$ along a straight track. (b) You walk for $1.00 \mathrm{~min}$ at a speed of $1.22 \mathrm{~m} / \mathrm{s}$ and then run for $1.00 \mathrm{~min}$ at $3.05 \mathrm{~m} / \mathrm{s}$ along a straight track. (c) Graph $x$ versus $t$ for both cases and indicate how the average velocity is found on the graph.

Compute your averagc velocity in the following two cascs:
(a) You walk $73.2 \mathrm{~m}$ at a speed of $1.22 \mathrm{~m} / \mathrm{s}$ and then run $73.2 \mathrm{~m}$ at a speed of $3.05 \mathrm{~m} / \mathrm{s}$ along a straight track. (b) You walk for $1.00 \mathrm{~min}$ at a speed of $1.22 \mathrm{~m} / \mathrm{s}$ and then run for $1.00 \mathrm{~min}$ at $3.05 \mathrm{~m} / \mathrm{s}$ along a straight track. (c) Graph $x$ versus $t$ for both cases and indicate how the average velocity is found on the graph.

Key Concepts

Average Velocity

Average velocity is defined as the total displacement divided by the total elapsed time. It provides an overall measure of how quickly an object moves from one position to another over an interval, regardless of the variations in speed during different segments of the journey.

Piecewise Motion

In situations where motion occurs in segments with distinct speeds or times, it is important to analyze each segment separately. For each segment, one computes the time or distance using the relationship between distance, speed, and time, and then combines these to determine the total displacement and total time.

Time and Distance Relationships

The fundamental relationship between distance, speed, and time (distance = speed × time) is central to solving motion problems. By applying this relationship to each segment of a piecewise motion, one can determine individual times when distances or speeds vary, which is necessary for computing overall quantities such as total time and average velocity.

Graphical Representation in Kinematics

A graph of position versus time provides a visual representation of motion. In these graphs, the slope at any point represents the instantaneous velocity, while the average velocity over a time interval is determined by the slope of the straight line (secant) connecting the starting and ending points of that interval. This graphical interpretation helps in understanding the overall motion characteristics.

Transcript

00:01 In this problem of motion along a straight line, we have to compute the average velocity in the following two cases.

00:09 So, we have given the first case.

00:12 In first case, you walk 73 .2 meters at a speed of 1 .22 meters per second.

00:20 So let's start.

00:21 So this is the origin from where we are starting our journey.

00:26 And it says that we are moving 73 .2 meters.

00:31 At a speed of 1 .22 meters per second.

00:34 So, say this is the point a and we have moved 73 .2 meters at the rate of, say, 1 .22 meters per second and then run 73 .2 meters at a speed of 3 .05 meters along the straight track.

00:57 So then again, we are moving to another point.

01:01 Point say b and the distance is again 73 .2 meters and the speed is here 3 .05 meters per second and in case b so this is the case b and in this case say this is again origin.

01:23 We are starting say you walk for 1 .00 minute at a speed of 1 .22 meters per second.

01:32 Now we are moving for 1 minutes.

01:35 So this is 1 minute and speed is 1 .22 meters per second.

01:40 1 .22 meters per second say this is the point a.

01:44 And then again then run for 1 minute.

01:49 Say then again we are running from here a to b.

01:52 So this is walk and this is for run.

01:56 Again for one minute at the speed of 3 .05 meter per so, this is 3 .05 meters per second.

02:07 And now we have to compute your average velocity in the following two cases.

02:13 So first case is here.

02:15 Now time taken say this is origin point o and here also this is origin point o.

02:21 Now time taken to move from zero or we can say the from point o2a.

02:28 So this value would be here.

02:30 We have the formula that is s is equals to v t now we want the time so t is calculated as distance that is s divided with the speed so here say this value would be t1 now this is 73 .2 meters divided with 1 .22 meters per second so this is 1 .22 meters per second so when we solve it this is equal to 60 seconds, this is 60 second.

03:05 Now for time t2 that means from e to b we are running.

03:09 So again here say this is t2.

03:11 Now we have the distance that is 73 .2 so this value would be 73 .2 using this formula that is time is equal to distance travel divided with the velocity with which we are traveling the distance.

03:26 Now velocity is 3 .05 so this value is 3 .05 meters per second.

03:33 Now when we solve it, t2 is equal to this is 24 seconds.

03:39 Now total time taken say this is t1 is equal to t1 plus t2 which is equals to 60 plus 24 that is 84 seconds.

03:51 Now we have to compare the average velocity.

03:54 So average velocity in case a.

03:58 So this value would be v average in case a.

04:01 Is equal to total distance or we can say this is in a straight line so this value would be total displacement that is 73 .2 plus 73 .2 which is equal to 146 .4 divided with time taken so time taken this is t1 is 60 and t2 is 24 for adding that we get the time that is equals to 84 seconds this is in meters now when we solve it this is equal to 1 .742 meters per second.

04:40 And now we have to solve it for case b.

04:44 So in case b, time is already mentioned...

Please give Ace some feedback