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

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எங்கள் ஆசிரியர்களுடன் 1-ஆன்-1 ஆலோசனை நேரத்தைப் பெறுங்கள். டாப்பர் ஆவதற்கு நாங்கள் பயிற்சி அளிப்போம்

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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: