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

hlo bro answer is easy hope u understand it=>>>

we know that
s = ut + (1/2)at
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/s .
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 = u(t-60) + (1/2).a.(t-60)
or as u = 0 and a = 0.2 m/s
s = (1/2)x0.2x(t-60)
so,
s = 0.1(t-60) ........................(1)
and
s = ut + (1/2)at
or
s = (1/2)x0.2xt
so,
s = 0.1t ........................(2)
now,
s = 3s
or
0.1(t-60) = 3x0.1t
or by taking square root of both sides
0.316.(t-60) = 0.547t
which can be calculated for 't'.


it it helps then brainliest pls

Answer 2

Answer:

2 min

Explanatiton:

let u for both the cars be =0

s1=ut+1/2at^2

=1/2(0.2)t^2

=0.1(t^2)-----------(2)

s2=1/2(0.2) (t-60)^2

 =t^2+120t+3600   ------------(1)

(1)=3*(2)