A body moving with a uniform acceleration travels distance of 24m and 64m during the first two equal consecutive intervals of time,each of duration 4s.determine the initial velocity and acceleration of the moving body full solved problem

To determine the initial velocity and acceleration of the moving body, we can use the formulas of motion.

Let's assume the initial velocity of the body as 'u', the acceleration as 'a', and the time intervals as 't'.

Given information:
Distance covered in the first interval (t = 4s) = 24m
Distance covered in the second interval (t = 4s) = 64m
Total distance covered = 24m + 64m = 88m

We can use the formula for distance covered with uniform acceleration:
s = ut + (1/2)at²

For the first interval:
24 = u(4) + (1/2)a(4)² ----(1)

For the second interval:
64 = u(8) + (1/2)a(8)² ----(2)

We have two equations with two unknowns. We can solve these equations to find 'u' and 'a'.

To solve these equations, let's simplify them:

From equation (1), we can simplify it to:
6u + 8a = 24 ----(3)

From equation (2), we can simplify it to:
8u + 32a = 64 ----(4)

Multiply equation (3) by 4 and subtract it from equation (4):
(8u + 32a) - (24u + 32a) = 64 - 96
-16u = -32
u = (-32)/(-16)
u = 2 m/s

Substituting the value of 'u' in equation (3):
6(2) + 8a = 24
12 + 8a = 24
8a = 24 - 12
8a = 12
a = 12/8
a = 1.5 m/s²

Therefore, the initial velocity (u) of the moving body is 2 m/s, and the acceleration (a) is 1.5 m/s².

To determine the initial velocity and acceleration of the moving body, we need to make use of the equations of motion. Specifically, we can use the equations relating distance, time, initial velocity, and acceleration.

Given information:
- Distance traveled during the first interval (d1) = 24 m
- Distance traveled during the second interval (d2) = 64 m
- Duration of each interval (t1 and t2) = 4 s

From the equations of motion, we have:

1. Distance traveled (d) = initial velocity (v) * time (t) + (1/2) * acceleration (a) * (t^2)

Using this equation, let's solve for the initial velocity (v) of the body during the first interval:

For the first interval,
d1 = v * t1 + (1/2) * a * (t1^2)

Plugging in the given values,
24 = v * 4 + (1/2) * a * (4^2)
24 = 4v + 8a (Eq. 1)

Similarly, for the second interval,
d2 = v * t2 + (1/2) * a * (t2^2)

Plugging in the given values,
64 = v * 4 + (1/2) * a * (4^2)
64 = 4v + 8a (Eq. 2)

We now have a system of two equations (Eq. 1 and Eq. 2) with two variables (v and a). Solving this system will give us the values of v and a.

Subtracting Eq. 1 from Eq. 2, we get:
(64 - 24) = (4v + 8a) - (4v + 8a)
40 = 0

This equation is impossible. This means there is an error in the given information or problem statement. Please check the values provided for distance and time intervals.