Question

using the graphical method derive the second equation of motions = ut - 1/2 at2​

Answer 1

Explanation:

pls mark as braileast answer

Answer 2

Explanation:

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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 ]