in this question I dont understand how he found the distances ??? Sa and Sb , and why he done that for Sb Sa = pi 88 why ???? why pi *88 Sb = pi 67 + 2 ( 88 - 67) ??why that please clarify for me that point
in this question I dont understand how he found the distances ??? Sa and Sb , and why he done that for Sb Sa = pi 88 why ???? why pi *88 Sb = pi 67 + 2 ( 88 - 67) ??why that please clarify for me that point
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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in this question I dont understand how he found the distances ??? Sa and Sb , and why he done that for Sb
Sa = pi 88 why ???? why pi *88
Sb = pi 67 + 2 ( 88 - 67) ??why that please clarify for me that point

Transcribed Image Text:Race car A follows path a-a while race car B follows path b-b on the unbanked track. If each car has a constant speed limited to that corresponding to a lateral (normal) acceleration of 0.91g, determine the times tą and tg for both cars to negotiate the turn as delimited by the
line C-C.
88 m/
67 m
a
Answers:
ta =
ts =
----

Transcribed Image Text:Lit Distance trauelled by ar A = SA
SA = T(88) = 276. 46 m.
%3D
et Distance trauclled by car B = Se
= TT(67) + 2 ( 88 -67)
%3D
= 252, 486m
SA
Time taben
Car A, tA =
&u .6רל
28.02
= 9,866 3
F 9,87 s (Ans)
Time taten by
SB
Car e = te =
Ve
252.486
24.456
= 10:324 E (Ans)
tA = 9.875
te = 10.324$
Pe, Car A wil win the race. (tating short time)
S6, Car A will win thc race.
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