a body moves in a straight line with initial velocity and uniform acceleration.find how far it travels in12s,given that it travel 246m in the 1st 6seconds,and 69m in the last 3seconds.find also the initial velocity.

To solve this problem, we can use the equations of motion for constant acceleration.

Let's break down the information given:
- The body travels 246m in the first 6 seconds.
- The body travels 69m in the last 3 seconds.
- We need to find the total distance traveled in 12 seconds and the initial velocity.

Step 1: Calculating the Acceleration
We know that acceleration is constant. Let's find the acceleration using the first 6 seconds:
Distance (d1) = 246m
Time (t1) = 6s
Using the equation: d1 = ut1 + (1/2)at1^2
Rearrange the equation to solve for acceleration (a):
a = (2(d1 - ut1)) / t1^2
Substituting the values:
a = (2(246 - u*6)) / 6^2
Simplifying:
a = (492 - 12u) / 36

Step 2: Calculating the Initial Velocity
We can now use the information in the last 3 seconds to find the initial velocity (u).
Distance (d2) = 69m
Time (t2) = 3s
Using the equation: d2 = ut2 + (1/2)at2^2
Substituting the values and the calculated value of acceleration (a):
69 = ut2 + (1/2)(492 - 12u)/36 * 3^2
Simplifying and solving for u:
69 = 3u + (246 - 6u)/12
Multiplying both sides by 12 to eliminate the fraction:
828 = 36u + 246 - 6u
Collecting like terms:
30u = 582
u = 582 / 30
u ≈ 19.4 m/s (approx.)

We have now found the initial velocity.

Step 3: Calculating the Total Distance
Now that we know the initial velocity (u) and acceleration (a), we can find the total distance (D) traveled in 12 seconds.
Using the equation: D = ut + (1/2)at^2
Total time (t) = 12s
D = 19.4 * 12 + (1/2)(492 - 12 * 19.4) / 36 * 12^2
Simplifying:
D = 232.8 + 24.5
D = 257.3 m

Therefore, the total distance traveled in 12 seconds is approximately 257.3 meters. The initial velocity is approximately 19.4 m/s.