Pls explain this question:

Two cars leave with a time gap of 1 minute from the same point. They move with an acceleration of 0.2 m/s2. How long after the departure of the second car, does the distance between them become equal to 3 times its value, when the second car just starts?

we know that

s = ut + (1/2)at²

This question states that from a point car A leaves and then after 1 minutes another car, car B leaves. Here we can assume that both cars start from rest and they have the same constant acceleration of 0.2 m/s².

now,

let the time at which teh distance travelled by A be three times that of B be 't - 60 secs' and thus time taken by B will be 't secs'

so,

s_A = u(t-60) + (1/2).a.(t-60)²

or as u = 0 and a = 0.2 m/s²

s_A = (1/2)x0.2x(t-60)² 

so,

s_A = 0.1(t-60)² ........................(1)

and

s_B = ut + (1/2)at²

or

s_B = (1/2)x0.2xt²

so,

s_B = 0.1t² ........................(2)

now,

s_A = 3s_B

or

0.1(t-60)² =  3x0.1t²

or by taking square root of both sides

0.316.(t-60) = 0.547t

which can be calculated for 't'.