a body is moving with a initial velocity of 2m/s with constant acceleration of2m/s^2.another body starts from rest with constant acceleration of 6m/s^2.find how far from the second particles overtake the first

x = 2 t + (1/2)(2) t^2

and
x = (1/2)(6) t^2
so
3 t^2 = 2 t + t^2
2 t^2 -2 t = 0
t (2 t - 2) = 0

t = 1
x = 3

To find the distance at which the second body overtakes the first body, we need to determine the time it takes for the second body to catch up to the first body.

Let's start by determining the time it takes for each body to reach the same velocity.

For the first body:
Initial velocity (u1) = 2 m/s
Acceleration (a1) = 2 m/s^2

Using the equation:
v1 = u1 + a1 * t

where v1 is the final velocity and t is the time, we substitute the known values:
0 (as v1 = 0 when the first body is overtaken) = 2 + 2 * t1

Solving for t1:
2 + 2 * t1 = 0
2 * t1 = -2
t1 = -1 second

Since time cannot be negative, we can ignore this negative solution. Thus, the first body reaches the velocity of the second body in 1 second.

Now, let's find the distance traveled by the second body during this time.

For the second body:
Initial velocity (u2) = 0 m/s (starts from rest)
Acceleration (a2) = 6 m/s^2

Using the equation:
s2 = u2 * t1 + 0.5 * a2 * t1^2

where s2 is the distance traveled by the second body, t1 is the time it takes to catch up, and u2 is the initial velocity of the second body.

Substituting the known values:
s2 = 0 * 1 + 0.5 * 6 * 1^2
s2 = 0 + 0.5 * 6 * 1
s2 = 0 + 3
s2 = 3 meters

Therefore, the distance from where the second body overtakes the first body is 3 meters.