Physics Challenge Question 2: Solutions
Question 1:
Time:
0.0 sec
Distance:
0.0 ft
Velocity:
1.0 sec
2.0 sec
3.0 sec
4.0 sec
5.0 sec
6.0 sec
5.0 ft
15 ft
25 ft
35 ft
55 ft
60 ft
5.0 ft/s
10 ft/s
10 ft/s
10 ft/s
20 ft/s
5.0 ft/s
Here, I just calculate the displacement and divide it by the time interval (each of those is 1 sec). For example, the
second interval is, with 2 significant figures,
15 ft  5.0 ft 10 ft
v

 10 fts
2.0s  1.0s 1.0s
Question 2:
The average velocity is
(5.0  10  10  10  20  5.0)
v average 
6
ft
s
 10
ft
s
Question 3:
I can use the first and last measurement to calculate the average velocity over the whole stretch. This is no
coincidence – what happens in between doesn’t matter for the average velocity!
60 ft  0 ft
v average 
 10 fts
6.0s  0.0s
Question 4:
1 ft = 0.3048 m, so vaverage  10 fts  0.3048 mft  3.0 ms