Question

Two cars leave with a time gap of 1 min from the same point. They move with an acceleration of 0.2m/s^2.how long after the departure of the second car does the distance between them become equal to three times its value , when the second car just starts

Answer 1

Answer:

It will take 1 min

Explanation:

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

Answer 2

Answer:

t=60s

Explanation:

car I drives 60s - S= ut+1/2at^2

                              = 1/2*0.2*60^2

                             = 360m

S-1= a(t+60)^2/2

S-2=at^2/2

S1-S2=a(t+60)^2-at^2/2

3*360=a[(t+60)^2-t^2)]/2

1080/a= t^2+60^2+2*t*60-t^2/2

1080/0.2= 3600+120t/2

5400= 1800+60t

5400-1800=60t

3600=60t

t=60s