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.
