Second and third equation of motion — lesson. Science CBSE, Class 9.

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Position-time relation:

Assume an object covering a distance (\(s\)) in time (\(t\)) in an uniform acceleration (\(a\)). The area enclosed within \(OABC\) gives the distance travelled by the object.

Velocity time graph

Distance (\(s\)) travelled by the object,

\begin{array}{l} s = area OABC (a trapezium) \\ = area of the rectangle OADC + area of the triangle ABD \\ = OA \times OC + \frac{1}{2} (AD \times BD) \end{array}

Substituting \(OA\) \(=\) \(u\), \(OC\) \(=\) \(AD\) \(=\) \(t\) and \(BD\) \(=\) \(at\), we get

\begin{array}{l} s = u \times t + \frac{1}{2} (t \times at) \\ s = ut + \frac{1}{2} a t^{2} \end{array}

Position-velocity relation:

The area enclosed within the trapezium \(OABC\) gives the distance (\(s\)) travelled by the object in time (\(t\)), moving under uniform acceleration (\(a\)).

\begin{array}{l} s = area of the trapezium OABC \\ s = \frac{(OA + BC) \times OC}{2} \end{array}

Substituting \(OA\) \(=\) \(u\), \(BC\) \(=\) \(v\) and \(OC\) \(=\) \(t\), we get
\(\)

s = \frac{(u + v) t}{2} \to (a)

From the velocity-time relation (from the second equation of motion),

t = \frac{v - u}{a} \to (b)

Substituting equation (b) on (a) we get,

s = \frac{(v + u) (v - u)}{2 a}

(or)

\begin{array}{l} 2 a s = v^{2} - u^{2} \\ v^{2} = u^{2} + 2 a s \end{array}

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14. Second and third equation of motion

Theory: