(a) as the second car has constant speed
we have
-4.2 + 2.70t = 5.60
t = 3.629 sec
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(b) both will have same speed
5.60 cm/s
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(c) equation for positions is given a
x1 = 14.5 - 4.2t + 1.35*t2
x2 = 9.5 + 5.6t
set x1 = x2
14.5 - 4.2t + 1.35*t2 = 9.5 + 5.6t
1.35t2 - 9.8t + 5 = 0
solve for t
we get
t = 9.8 +/- 8.31 / 2.7
t = 6.71 sec and t = 0.5518 sec
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(d) locations where they pass one another
we get
x1 = 9.5 + 5.6t = 12.6 cm
x2 = 9.5 + 5.6t = 47.1 cm
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