Question

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 briskly in a straight line at a constant speed of 2.20 m/s. What are (a) his average velocity vavg and (b) his average acceleration aavg in the time interval 2.00 min to 8.00 min? What are (c) vavg and (d) aavg in the time interval 3.00 min to 9.00 min? (e) Sketch x versus t and v versus t, and indicate how the answers to (a) through (d) can be obtained from the graphs.

Answer 1

So the question ask to calculate the following base on the given data in your problem.
a. average velocity is 66m/min
b. average acceleration is 22m/min^2
c. The time interval of average velocity 88m/min
d. The average acceleration is 22m/m^2

I hope you are satisfied with my answer and feel free to ask for more

Answer 2

1) Definitions

Average velocity = change in position / total time

Average acceleration = change in velocity / total time

2) (a) Average velocity in the time interval 2.00min to 8.00 min

i) time elapsed = 8.00min - 2.00min = 6.00 min = 360 s

ii) initial position at t = 2.00: x = 0 (the man remained still until t = 5.00)

iii) final position at t = 8.00:

constant speed from t = 5.00 to t = 8.00, V = 2.20 m/s

Change in position = constant velocity × time =

time = 8.00min - 5.00min = 3.00 min = 180s

Change in position = 2.20 m/s × 180 s = 396m

iv) Compute the verage velocity

Average velocity = change in position / time elapsed = 396m / 360 s = 1.10 m/s

3) (b) Average acceleration in the time interval 2.00 min to 8.00 min

i) time elapsed: 360 s

ii) initial velocity, at t = 2.00 min: 0

iii) final velocity, at t = 8.00 min: 2.20 m/s

iv) Compute the average acceleration

Average acceleration = change in velocity / time elapsed = 2.20 m/s / 360s = 0.00611 m/s²

4) (c) Average velocity in the time interval 3.00min to 9.00min

i) time elapsed: 9.00min - 3.00min = 6.00 min = 360 s

ii) initial position, at t = 3.00 min: 0

iii) final position, at t = 9.00 min

change in position = constant speed × time

time = 9.00min - 5.00min = 4.00 min = 240s

displacement = 2.20m/s × 240s = 528m

iv) Compute the average velocity

Average velocity = 528m / 360 s = 1.47 m/s

5) (d) Average acceleration in the interval 3.00 min to 9.00 min?

i) change in velocity: 2.20 m/s

ii) time elapsed = 6.00min = 360 s

iii) compute: 2.20m/s / 360s = 0.00611 m/s²

6 (e) Sketch x versus t and v versus t, and indicate how the answers to (a) through (d) can be obtained from the graphs.

i) to do the sketches use these tables

x versus time

t -------- x

0 ------- 0

2 ------- 0

3 ------- 0

5 ------- 0

8 ------- 396

9 ------ 528

10 ----- 660

velocity versus time

t ------- v

0 ----- 0

2 ----- 0

3 ----- 0

5 ----- 0

8 ----- 2.20

9 ----- 2.20

10 --- 2.20

Here the times are in minutes and the speeds in m/s

ii) You can obtain the answers to (a) through (d) by using these facts:

- speed = slope of the graph x versus t

- acceleration = slope of the graph v versust t.

Then, for each case, take the extremes of the time intervals and find the quotient (divide) rise / run, i.e. vertical change / horizontal change, between the extreme points of your time interval.

Doing that for the graph x versus t you obtaind the average speed, and for the graph v versus t you obtain average acceleration.