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