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Question

Q.Two cars leave with a one minute gap and move with an accelaration of 0.2m/s2.How long after the departure of second car the distance between them become equal to three times its initial value?

a.2 minute

b.3/2 minute

c.1 minute

d.1/2 minute

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Solution

car1 start from rest and drives 60 s => s = at²/2 = 0.5*60²/2 = 900 m

car 2 starts when car 1 has driven 900 m.

car 1 drives s1 = a(t + 60)²/2

car 2 drives s2 = at²/2

The diffence s1 - s2 = 3* 900 = 2700

a(t+60)²/2 - at²/2 = 2700
(t+60)² - t² = 5400/a
120t + 3600 = 5400/0.5 = 10800
t = (10800-3600)/120 = 60 s

6s = 1 minute

option C is correct

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