Solve Time Interval Between Two Cars Leaving with 0.4 m/s^2

[sambarbarian]

Two cars leave one after the other and travel with acceleration 0.4 m/s^2 . Two minutes after the departure of the first vehicle , the distance between the cars becomes 1.9 km . The time interval between the departure of the cars is ?



i tried to do this by :- for First car (A)... 0.4 = Final V - 0 /120

therefore Final V is 48 m/s .

average V = 48 - 0 / 2 = 24

and 24 = d / 120 ... therefore distance = 24 * 120 .

now 24 * 120 - 1900 is the distance traveled by second car ( B ) ..

how should i proceed , if my logic is correct till now ?

 

[Simon Bridge]

You want kinematics ... look up the suvat equations and work in terms of distance traveled.

Break it down:
In the first 2min, how far has the first car travelled?
(You can do this directly from the time and the acceleration without referring to final velocity... but you are fine.)

How far has the second car travelled? (you've got this.)

How long has the second car been traveling to get there?
(Here is where you want that suvat equation then - you know s, u, and a; you want t.)

The time interval between departures.

 

[sambarbarian]

u mean S = ut + at^2/ 2 ??ok i got it , thank you .

 

[Simon Bridge]

well done :)