Chapter 2-1 pg 69 Review ANSWERS
1) The distance traveled is 6 m, the displacement is also 6 m
2) On a position time graph the slope of the curve at any point represents the velocity.
3) An object at rest will have a horizontal line (straight, with a zero slope)
An object with constant positive velocity will have a diagonal line on a position time graph, going up to the right.
An object with constant negative velocity will have a diagonal line on a position time graph going down to the right. (negative slope)
4) For the graph shown in Fig 2-19 on pg 60 for a bug crawling along a line (ewwww!),
the velocity is positive from t2 to t4, at t1 it is negative, at t2 it is positive, at t3 it is positive, at t4 it is zero, at t5 it is zero again.
5) the VELOCITY is increasing from t1 to t3, and from t4 and a half to t5,
b) the VELOCITY is decreasing from 0 to t1, from t3 to t4 and a half.
Note…. The SPEED is increasing from t1 and a half to t3, t4 to t4 and a half…
6) if the Vavg is 0 then the displacement is zero as well.
7) delta t is always positive because time always goes forward (as far as YOU know!)
8) Vavg = D/T = 1142 km/15.08 hr = 75.73 km/hr
9) Vavg = D /T = 10km/ .53 hr = 18.87 km/hr
10) for the position time graph for fig 2-20 on page 69,
a) for t=0 to t = 3 sec, the squirrel’s displacement is -2 m.
b) the Vavg is D/T = -2/3 = -.67 m/s
11) Vavg = D/t = 42.19 km/ (2h, 9 min, 21 s)
2 h * 3600 s + 9 min * 60 sec + 21 s = 7761 sec
1 hr 1 min
42.19 km = 42190 m
Vavg= 42190 m/ 7761 sec = 5.436 m/s