Question

Derive first equation of motion v=u + at and second equation of motion s=ut+1/2 at2 using graphical method.

Answer 1

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$\underline{ \bigstar \: \textsf{First Equation of Motion .}}$

Let us consider an object is moving with uniform acceleration along a straight line . Let (u) be the initial velocity at t = 0 . After interval of time (t) , its velocity becomes u.

➳ Let OA = u [ t = 0 ]

Here in the figure we draw AD ⟂ BC , and BE ⟂ OY.

➳ ∠BAD = ϴ. [ say ]

Now,

➳ Acceleration = slope of Line AB

➳ a = tan ϴ

➳ a = BD/AD. [ tan ϴ = P/B ]

➳ a = (v - u)/t. [ AD = OC = t ]

or, at = v - u.

or, v = u + at. [ 1st Equation of motion ]

_______________…

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$\\ \underline{ \bigstar \: \textsf{Second Equation of Motion .}}$

Let us consider, an object is moving along a straight line with a uniform acceleration (a) at t = 0 , initial velocity (u) . Let s be the distance travelled by the object in time t , from A to b.

➳ s = area enclosed by velocity time graph and time axis.

➳ s = Area of OABC

➳ s = Area of OACD + Area of ∆ ABD.

➳ s = OA × AD + 1/2 × BD × AD

➳ s = OA × OC + 1/2 (BC - CD) × OC. ...Eq(1)

• OA = u
• OC = t
• BD = BC - CD = v - u.

➳ s = u × t + 1/2 (v - u) t

From first equation of motion we have , v - u = at.

➳ s = u × t × 1/2 × at × t

➳ s = ut + 1/2at². [ 2nd equation of motion ]

Answer 2

$\huge \fbox \green{1st\: equation\:of\:motion}$

• initial velocity (u) = OA
• final velocity (v) = BC
• acceleration (a) = Slope of the graph

Refer to the attachment for the Graph.

According to the diagram:-

$\longrightarrow$ acceleration = $\bf\dfrac{BD}{AD}$

$\longrightarrow$ a = $\bf \dfrac{BC-CD}{OC}$ (since, AD = OC)

$\longrightarrow$ a = $\sf \dfrac{v-u}{t}$

$\longrightarrow$ at = v-u

$\longrightarrow$ $\underline{\boxed{\sf v = u+at }}$

$\huge \fbox \green{2nd\: equation\:of\:motion}$

• velocity = $\sf\dfrac{Displacement}{time}$
• Displacement = velocity × time
• Displacement = average velocity × time

Displacement = $\sf\dfrac{initial\: velocity+final\: velocity}{2}$ × time

Displacement (S) = $\sf\dfrac{v+u}{2}$ × t ... $\sf eq_{1}$

we know that, v = u+at

So, putting the value of v in $\sf eq_{1}$

$\longrightarrow$ S = $\sf\dfrac{u+at+u}{2}$ × t

$\longrightarrow$ S = $\sf\dfrac{2u+at}{2}$ × t

$\longrightarrow$ S = $\sf\dfrac{2u}{2}$ + $\sf\dfrac{at}{2}$ × t

$\longrightarrow$ S = $\sf ut + \dfrac{at²}{2}$

$\longrightarrow$ $\underline{\boxed{\sf S = ut+\dfrac{1}{2}at²}}$

Additional information:-

3rd equation of motion:-

$\longrightarrow$ $\underline{\boxed{\sf v²=u²+2as}}$