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**Hint:**If the body is moving at a constant acceleration and doesn’t change directions during the nth second, then we can use some special equations for kinematics to calculate the “displacement” which will equal the “distance travelled” only if the direction does not change and the formula is:

${S_n} = u + \dfrac{a}{2}(2n - 1)$

Where,

u = Initial velocity of the object.

a = Initial acceleration of the object.

**Complete step by step answer:**

We know that distance travelled in the nth second is given by:

${S_n} = u + \dfrac{a}{2}(2n - 1)$

Where,

u = Initial velocity of the object.

a = Acceleration of the particle.

It is given that at \[{5^{th}}\] second the object travelled 22 m.

Then,

$ \Rightarrow {S_5} = u + \dfrac{a}{2}(2\left( 5 \right) - 1)$

\[ \Rightarrow 25 = u + \dfrac{{9a}}{2}\] … (i)

Similarly, the distance travelled in \[{7^{th}}\] second is 33 m. Then,

$ \Rightarrow {S_7} = u + \dfrac{a}{2}(2\left( 7 \right) - 1)$

$ \Rightarrow 33 = u + \dfrac{{13a}}{2}$ … (2)

On solving equation (1) and equation (2):

Subtracting eq. (1) in eq. (2) i.e., (2) – (1)

$ \Rightarrow 33 - 25 = \dfrac{a}{2}\left( {13 - 9} \right)$

$ \Rightarrow 8 = \dfrac{a}{2}\left( 4 \right)$

$ \Rightarrow a = 4\dfrac{m}{{{s^2}}}$

Therefore, the particle was moving an initial acceleration of $4\dfrac{m}{{{s^2}}}$

Substituting value of ‘a’ in equation (1) and simplifying:

\[ \Rightarrow 25 = u + \dfrac{{9 \times 4}}{2}\]

\[ \Rightarrow u = 25 - 18 = 7\dfrac{m}{s}\]

$ \Rightarrow u = 7\dfrac{m}{s}$

Therefore, the particle was moving with an initial velocity of $u = 7\dfrac{m}{s}$ .

Hence, initial velocity and acceleration of the particle is $7\dfrac{m}{s}$ and $4\dfrac{m}{{{s^2}}}$ .

**Note:**

One should remember that the above approach is valid if and only if the object is moving with “constant acceleration” and the displacement is taking place along the “same direction”. If the direction of displacements are different then we cannot proceed with this method.

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