**Unformatted text preview: **traight portion
of the curve. Use a least-squares ﬁt to determine this
slope.
t (s) 20.0°
T mg Figure P6.69
70. A 9.00-kg object starting from rest falls through a viscous medium and experiences a resistive force R
b v, where v is the velocity of the object. If the object’s
speed reaches one-half its terminal speed in 5.54 s,
(a) determine the terminal speed. (b) At what time is
the speed of the object three-fourths the terminal
speed? (c) How far has the object traveled in the ﬁrst
5.54 s of motion?
71. Members of a skydiving club were given the following
data to use in planning their jumps. In the table, d is
the distance fallen from rest by a sky diver in a “free-fall d (ft) 1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20 16
62
138
242
366
504
652
808
971
1 138
1 309
1 483
1 657
1 831
2 005
2 179
2 353
2 527
2 701
2 875 ANSWERS TO QUICK QUIZZES
6.1 No. The tangential acceleration changes just the speed
part of the velocity vector. For the car to move in a circle, the direction of its velocity vector must change, and
the only way this can happen is for there to be a centripetal acceleration.
6.2 (a) The ball travels in a circular path that has a larger radius than the original circular path, and so there must
be some external force causing the change in the velocity vector’s direction. The external force must not be as
strong as the original tension in the string because if it
were, the ball would follow the original path. (b) The
ball again travels in an arc, implying some kind of external force. As in part (a), the external force is directed toward the center of the new arc and not toward the center of the original circular path. (c) The ball undergoes
an abrupt change in velocity — from tangent to the circle to perpendicular to it — and so must have experienced a large force that had one component opposite
the ball’s velocity (tangent to the circle) and another
component radially outward. (d) The ball travels in a
straight line tangent to the original path. If there is an
external force, it cannot have a component perpendicular to this line because if it did, the path would curve. In fact, if the string breaks and there is no other force acting on the ball, Newton’s ﬁrst law says the ball will travel
along such a tangent line at constant speed.
6.3 At the path is along the circumference of the larger
circle. Therefore, the wire must be exerting a force on
the bead directed toward the center of the circle. Because the speed is constant, there is no tangential force
component. At the path is not curved, and so the wire
exerts no force on the bead. At the path is again
curved, and so the wire is again exerting a force on the
bead. This time the force is directed toward the center
of the smaller circle. Because the radius of this circle is
smaller, the magnitude of the force exerted on the bead
is larger here than at ....

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