From t=0 to t=5.00 min, a man stands still, and from t=5.00 min to t=10.0 min, he walks brisk

From t=0 to t=5.00 min, a man stands still, and from t=5.00...

From $t=0$ to $t=5.00 \mathrm{~min},$ a man stands still, and from $t=5.00 \mathrm{~min}$ to $t=10.0 \mathrm{~min},$ he walks briskly in a straight line at a constant speed of $2.20 \mathrm{~m} / \mathrm{s}$. What are (a) his average velocity $v_{\text {arg }}$ and (b) his average acceleration $a_{\mathrm{my}}$ in the time interval $2.00 \mathrm{~min}$ to to 9.00 min? (e) Sketch $x$ versus $f$ and $v$ versus $t,$ and indicate how the answers to (a) through (d) can be obtained from the graphs.

From $t=0$ to $t=5.00 \mathrm{~min},$ a man stands still, and from $t=5.00 \mathrm{~min}$ to $t=10.0 \mathrm{~min},$ he walks briskly in a straight line at a constant speed of $2.20 \mathrm{~m} / \mathrm{s}$. What are (a) his average velocity $v_{\text {arg }}$ and (b) his average acceleration $a_{\mathrm{my}}$ in the time interval $2.00 \mathrm{~min}$ to
to 9.00 min? (e) Sketch $x$ versus $f$ and $v$ versus $t,$ and indicate how the answers to (a) through (d) can be obtained from the graphs.

Key Concepts

Displacement

Displacement represents the change in position of an object from its initial point to its final point, considering both magnitude and direction. It is the straight-line distance between these two positions and is crucial for calculating average velocity.

Graphical Analysis in Kinematics

Graphical analysis in kinematics involves the use of plots such as displacement versus time and velocity versus time graphs to analyze motion. The slope of a displacement-time graph corresponds to velocity, while the slope of a velocity-time graph corresponds to acceleration. These graphs are essential tools for visually interpreting and deriving motion parameters.

Average Velocity

Average velocity is defined as the net displacement divided by the total time taken. It is a vector quantity that considers both the magnitude and the direction of the movement, representing how fast something has moved overall, rather than the instantaneous speed at any moment.

Average Acceleration

Average acceleration is the change in velocity over a given time interval. It quantifies how quickly an object’s velocity increases or decreases and is a fundamental concept in kinematics, particularly when analyzing changes in the motion such as speeding up, slowing down, or changing direction.

Transcript

00:01 In this problem of motion in a straight line we have given that from t is equal to 0 to t is equals to 5 minutes so t is equals to 5 minutes a man stands still and from t is equals to 5 minute to t is equal to so here man is in still and now from t is equal to 5 minute to t is equals to 10 minutes he walks briskly in a straight line with a constant speed of v is equal to 2 .20 meter per second.

00:45 Now what are his average velocity? first we have to find the average velocity.

00:52 His average acceleration in the time interval two minute to eight minutes.

00:58 So from here we have to find the time interval which is eight minute to two minutes which is equals to 6 minutes which is equivalent to 360 seconds so we can say this is equals to 360 seconds whereas the sub interval in which he is moving is about say 5 to 10 minutes so we have only we say that this is delta t this is only for 3 minutes because because from 0 to 5 minute he is in still from 5 to 10 minutes he is working so we have only 3 minutes of movement so which is equals to 180 seconds now at t is equal to 2 minute the man was at rest so we can say at t is equals to 0 minute so we can say t is equal to 0 minute the particle was not moving that is it is a 0 position and when it was at 8 minutes or we can say here here t is equals to t minutes or we can say two minutes the position was at rest now t is equals to eight minutes we have to calculate the position change so we can say here x is equals to now we have moved we have to only 180 second with with the above 2 .20 meter per second so we have 396 meters so at t is equal to eight minutes the man has moved 396 meters from the initial position now we have to calculate average velocity is change in position so change in position is 396 meter with the time so time has been taken that is 360 seconds so when we calculate this is equals to 1 .10 meter per second now for part b so here this is part b part b says his average acceleration in the time interval two minutes to to 8 minute.

03:14 Now we have to calculate average acceleration.

03:16 So a average acceleration.

03:19 Now it has moved to 2 .2 meter per second from 0.

03:26 So we say that this is 2 .20 meters per second minus 0 and divided with the total time taken which is 360 seconds.

03:37 So here we have meter per second square.

03:40 So average acceleration is equals to this is 0 .00611 meters per second square.

03:52 So this is the answer for part b.

03:54 Now part c...