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To calculate the initial velocity and acceleration of an object moving with uniform acceleration, we can use the kinematic equation:

\[s = ut + \frac{1}{2}at^2\]

where:

- \(s\) is the displacement

- \(u\) is the initial velocity

- \(t\) is the time

- \(a\) is the acceleration

We have two sets of data:

1. Displacement of 25 meters in 5 seconds: \(s_1 = 25\) meters, \(t_1 = 5\) seconds

2. Displacement of 36 meters in 6 seconds: \(s_2 = 36\) meters, \(t_2 = 6\) seconds

Let's calculate the acceleration first:

For the first set of data:

\[s_1 = ut_1 + \frac{1}{2}at_1^2\]

\[25 = 5u + \frac{1}{2}a(5^2)\]

\[25 = 5u + 12.5a \quad \text{(equation 1)}\]

For the second set of data:

\[s_2 = ut_2 + \frac{1}{2}at_2^2\]

\[36 = 6u + \frac{1}{2}a(6^2)\]

\[36 = 6u + 18a \quad \text{(equation 2)}\]

To solve these two equations simultaneously, we can eliminate \(u\) by multiplying equation 1 by 6 and equation 2 by 5:

\[150 = 30u + 75a\]

\[180 = 30u + 90a\]

Subtracting the first equation from the second equation, we get:

\[180 - 150 = 90a - 75a\]

\[30 = 15a\]

\[a = 2 \, \text{m/s}^2\]

Now we can substitute the value of \(a\) into equation 1 to find \(u\):

\[25 = 5u + 12.5(2)\]

\[25 = 5u + 25\]

\[5u = 25 - 25\]

\[5u = 0\]

\[u = 0 \, \text{m/s}\]

Therefore, the initial velocity is \(0 \, \text{m/s}\) and the acceleration is \(2 \, \text{m/s}^2\).

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V=u +at

After 5s

25 = u + 5a ——(1)

After 8s

34 = u + 8a ——(2)

Subtract equation (1) from equation (2)

34 - 25 = u + 8a - u - 5a

9 = 3a

a = 3 m/s^2

Substitute a = 3m/s^2 in equation (1)

25 = u + 5*3

u = 10 m/s

Distance travelled in 12th second

S = u + a(n - 1/2)

S = 10 + 3*(12 - 1/2)

S = 44.5m

Distance travelled in 12th second is 44.5m

Hope it is helpful to u