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???????
Please give the answer

Answer 1

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,

sA = u(t-60) + (1/2).a.(t-60)2

or as u = 0 and a = 0.2 m/s2

sA = (1/2)x0.2x(t-60)2

so,

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

and

sB = ut + (1/2)at2

or

sB = (1/2)x0.2xt2

so,

sB = 0.1t2 ........................(2)

now,

sA = 3sB

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'