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.

V = Vo + a t

X = Vo*t + (a/2) t^2 (distance travelled at time t)
Vo is the initial velocity and a is the acceleration.

Velocity at t = 10.5 s = 69/3 = 23 m/s
Velocity at t = 3 s = 246/6 = 41 m/s
Acceleration = (23 - 41)/7.5 = -2.4 m/s^2

41 = Vo + 3a = Vo - 7.2
Vo = 48.2 m/s

Total distance travelled at t = 12 s
= 48.2*12 - 1.2*144 = 405.6 m

thanks alot

To solve this problem, we can use the kinematic equation:

distance (d) = initial velocity (u) * time (t) + (1/2) * acceleration (a) * time (t)^2

We are given specific distances for the first 6 seconds and the last 3 seconds, which can help us find the acceleration (a). Here's how we do it step by step:

Step 1: Calculate the acceleration (a):
Using the distance formula, d = ut + (1/2)at^2, for the first 6 seconds:
246 = u * 6 + (1/2) * a * 6^2
246 = 6u + 18a (equation 1)

Using the distance formula, d = ut + (1/2)at^2, for the last 3 seconds:
69 = u * 3 + (1/2) * a * 3^2
69 = 3u + 4.5a (equation 2)

Step 2: Solve the simultaneous equations:
From equation 1, we can express 6u as 246 - 18a and substitute it into equation 2:
69 = (246 - 18a) + 4.5a
69 = 246 - 18a + 4.5a
69 - 246 = -13.5a
-177 = -13.5a
a = (-177) / (-13.5) = 13.11 m/s^2 (approx.)

Step 3: Find the initial velocity (u):
Substitute the value of acceleration (a) into equation 1:
246 = 6u + 18 * 13.11
246 = 6u + 235.98
6u = 246 - 235.98
6u = 10.02
u = 10.02 / 6 = 1.67 m/s (approx.)

Step 4: Calculate the total distance traveled in 12 seconds:
Using the distance formula for the entire 12 seconds:
d = ut + (1/2)at^2
d = 1.67 * 12 + (1/2) * 13.11 * 12^2
d = 20.04 + 93.852
d = 113.892 m (approx.)

Therefore, the body travels approximately 113.892 meters in 12 seconds, and the initial velocity is approximately 1.67 m/s.