Chapter 2 P 9 Physics 210 Problems - My Solutions
Dr. Hulan E. Jack Jr.
Serway, Faughn and Vuille: College Physics 8th Edition , Thomson Brooks/Cole, Vol I(ISBN #) 978-049511374-3
THE PROBLEM STATEMENT
Ch 2 9. Find the instantaneous velocities of the
tennis player of Figure P2.8 at
(a) 0.50 s,
(b) 2.0 s,
(c) 3.0 s, and
(d) 4.5 s.
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Physics 210 Problems - My Solutions
Dr. Hulan E. Jack Jr.
Ch 2 9. Find the instantaneous velocities of the
tennis player of Figure P2.8 at
(a) 0.50 s,
(b) 2.0 s,
(c) 3.0 s, and
(d) 4.5 s.
BRAINSTORMING-Definitions,
concepts , principles and Discussion
The instantaneous is the x/t over an ever decreasingly small time interval t.
It is the slope of the x vs. t during that decreasing time interval. Here we have the situation where
the slope in constant over an extended, a finite, interval. So the instantaneous velocity is the
same for every point in that interval.
For example, the instantaneous velocity of every point in the time interval (0.0s,1.0s) from 0.0s
to 1.0s , is the same,
x(t=1s) = 4m x(t=0s) = 0m
vinstant = x/t(over the interval) = (x(t=1s) - x(t=0s))m/(1s-0s) = (4m-0m)/1s = 4m/1s =4 m/s.
t = 0.5s is in this interval, So,
vinstant (0.50s) = 4 m/s .
The same definition applies for all of the intervals.
Page 2 of 3
Physics 210 Problems - My Solutions
Dr. Hulan E. Jack Jr.
Ch 2 9. Find the instantaneous velocities of the
tennis player of Figure P2.8 at
(a) 0.50 s,
(b) 2.0 s,
(c) 3.0 s, and
(d) 4.5 s.
Basic Solution
Since the slope of the x vs t curve is
constant over extended intervals, the instantaneous velocity is constant over each
interval and equals the slope of that interval. So,
0.5s lies between 0s and 1s, and
x(t=1s) = 4m x(t=0s) = 0m.
vinstant (0.50s) = x/t over the (1.0s,0.0s) time interval = 4m/1s = 4 m/s.
2.0s lies between 1s and 2.5s,
and x(t=2.5s) = -2m x(t=1.0s) = 4m.
vinstant (2.0s) = x/t over the (4.0s,2.5s) time interval = (-2-4)m/(2.5,1.0)s = -6m/1.5s = -4 m/s.
3.0s lies between 2.5s and 4s,
and x(t=4.0s) = -2m x(t=2.50s) = -2m.
vinstant (3.0s) = x/t over the (4.0s,2.5s) time interval = (-2-(-2))m/(4.0,2.5)s = (-2+2)m/1.5s
= 0 m/1.5s = 0 m/s.
4.5s lies between 4s and 5 s,
and x(t=5.0s) = 0m x(t=4.0s) = -2m.
vinstant (4.5s) = x/t over the (5.0s,4.0s) time interval = (0-(-2))m/(5.0,4.0)s = (0+2)m/1s =
2m/1s = 2 m/s.
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