Question

two cars leave with the time gap of for 1 minute from the same point with an acceleration of 0.2 metre per second square how long after the departure of the second is the distance between them becomes equal to 3 times its value when second car just starts​

Answer 1

Answer:

Hey!

Distance covered by car in 1 st

1 min=21 ×0.2×60×60

=360 m

velocity of 1 min

v=12 m/s

O

rest = 0 O vel

=12m/s Road S vel

=3×360m=1080m

t = ?

S rest = 21a rest 2 +O rest.

= 21×0×t 2 +O rest.1080=12×t.

t=90sec

=> 1.5min

Hope it will be helpful ✌️

Answer 2

Explanation:

Aloha!

s¹=a(t+60)²/2

s²=a(t)²/2

S=s¹-s²=2700

=a(t+60)²/2-a(t)²/2=2700

=a(t+60)²/2-at²/2=2700

=(t+60)²-t²=5400/a

=t²+3600+120t-t²=5400/a

=120t+3600=5400/0.5

=120t+3600=10800

=120t=7200

=t=60

Thank you

@ Twilight Astro ✌️☺️❤️