p02 084

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84. We note that the running time for Bill Rodgers is ∆t

R

= 2(3600) + 10(60) = 7800 s. We also note that

the magnitude of the average velocity (Eq. 2-2) and Eq. 2-3 (for average speed) agree in this exercise
(which is not usually the case).

(a) Denoting the Lewis’ average velocity as v

L

(similarly for Rodgers), we find

v

L

=

100 m

10 s

= 10 m/s

v

R

=

42000 m

7800 s

= 5.4 m/s .

(b) If Lewis continued at this rate, he would covered D = 42000 m in

t

L

=

D

v

L

=

42000

10

= 4200 s

which is equivalent to 1 h and 10 min.


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