RECITATION PROBLEMS
CHAPTER 3
WEEK 4
3-3
A dragonfly flies from point A to point B along the path shown in the following figure in
1.50 s. (a) Find the x and y components of its position vector at point A. (b) What are
the magnitude and direction of its position vector at A? (c) Find the x and y components
of the dragonfly’s average velocity between A and B. (d) What are the magnitude and
direction of its average velocity between these two points?
Coordinates for point A are (2.0 m, 1.0 m) and for point B they are (10.0 m, 6.0 m).
(a) At A, x = 2.0 m, y = 1.0 m.

(b) The magnitude of position vector r is r  x 2  y 2  2.2m and
 y
x
 1.0m 
  26.6 counterclockwise from the +x axis.
 2.0m 
  tan 1    tan 1 
(c) The x component of the dragonfly’s average velocity between A and B:
x 10.0m  2.0m
vav, x 

 5.3m / s
t
1.50 s
y 6.0m  1.0m
vav, y 

 3.3m / s
t
1.50 s
(d) The magnitude of average velocity between A and B:
vav 
v   v 
2
av, x
2
av, y

5.3m / s 2  3.3m / s 2
1
 6.2m / s
The direction of average velocity vector:
 vav, x 
  tan 1  3.3m / s   32 counterclockwise from the +x axis.

 5.3m / s 
 vav, y 
  tan 1 
3-5
An athlete starts at point A and runs at a constant speed of 6.0 m/s around a round track
100 m in diameter, as shown in the following figure. Find the x and y components of this
runner’s average velocity and average acceleration between points (a) A and B, (b) A and
C, (c) C and D, and (d) A and A (full lap). (e) Calculate the magnitude of the runner’s
average velocity between A and B. Is his average speed equal to the magnitude of his
average velocity. Why or why not? (f) How can his velocity be changing if he is
running at constant speed?
(a)
The time for one full lap is
2r 2 50m 
t

 52.4 s.
v
6.0m / s
A to B is one-quarter lap and takes
x 0   50m 

 3 . 8m / s
t
13.1s
y  50m  0


 3 .8m / s
t
13.1s
1
52.4s   13.1s.
4
vav, x 
vav, y
v x 6.0m / s  0

 0.46m / s 2
t
13.1s
v y 0  6.0m / s


 0.46m / s 2
t
13.1s
aav, x 
aav, y
2
(b)
The time for A to C (half of the lap) is
1
t  52.4 s   26.2 s
2
X and y components for the average velocity and average acceleration are as follows:
x  50m   50m 

 3.8m / s
t
26.2 s
y

0
t
vav, x 
vav, y
v x
0
t
v y  6.0m / s  6.0m / s


 0.46m / s 2
t
26.2 s
aav, x 
aav, y
(c)
The time for C to D is
1
t  52.4s   13.1s
4
x 0  50m

 3.8m / s
t
13.1s
y  50m  0


 3.8m / s
t
13.1s
v av, x 
v av, y
v x  6.0m / s  0

 0.46m / s 2
t
13.1s
v y 0   6.0m / s 


 0.46m / s 2
t
13.1s
a av, x 
a av, y
(d)
For A to A (full lap):
x  y  0 so v av, x  v av, y  0 , and v x  v y  0 so a av, x  a av, y  0
(e)
2
2
For A to B, v av  v av
, x  v av, x 
3.8m / s 2  3.8m / s 2
3
 5.4m / s .
Since the speed is constant, the average speed is 6.0 m/s. The distance traveled is larger
than the displacement and hence the average speed is larger than the magnitude of the
average velocity.
(f)
Here, the direction of the velocity vector is changing. Velocity is a vector quantity. Any
change in its magnitude or direction or both causes acceleration.
Here, acceleration describes the rate of change of direction of the velocity.
3-31
You swing a 2.2 kg stone in a circle of radius 75 cm. At what speed should you swing it
so its centripetal acceleration will be 9.8 m/s2?
Radius R = 75 cm = 0.75 m, centripetal acceleration a rad  9.8m / s 2 .
a rad 
v2
and hence v  Ra rad 
R
0.75m9.8m / s 2   2.7m / s .
3-35
A wall clock has a second hand 15.0 cm long. What is the radial acceleration of the tip of
this hand?
Radius R=0.150 m, T = 1 min = 60.0 s.
2
2R
2
2
2
v
T  4 R  4 0.150m   1.643 10 3 m / s 2
arad 

R
R
T2
60.0s 2


3-37
Dizziness. Our balance is maintained, at least in part, by the endolymph fluid in the inner
ear. Spinning displaces this fluid, causing dizziness. Suppose a dancer (or skater) is
spinning at a very high 3.0 revolutions per second about a vertical axis through the center
of his head. Although the distance varies from person to person, the inner ear is
approximately 7.0 cm from the axis of spin. What is the radial acceleration (in m/s2 and
in g’s) of the endolymphfluid?
R = 0.070 m.
For 3.0 rev/s, the period T is T 
2R T 

1 .0 s
 0.333s
3.0rev
2
v
4 2 R 4 2 0.070m 


 25m / s 2  2.5 g
2
2
R
R
T
0.333s 
The force on the fluid must be 2.5 times its weight.
arad 
2
4