Question

derive v^2=u^2+2as using v-t graph
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Answer 1

To proof :

• v² = u² + 2as

Prove :

→ v² = u² + 2as

Where,

• v = Final velocity
• u = Initial velocity
• a = Acceleration
• s = Distance

Now,

→ s = Area of trapezium OABCD

→ s = ½ (sum of parallel side) × height

→ s = ½ × (OA + BD) × OD

→ s = ½ × (v + u) × t

→ s = ½ (v + u) × (v - u/a)

°.° t = v - u/a

→ s = v² - u²/2a

→ 2as = v² - u²

→ v² - u² = 2as

→ v² = u² + 2as ......(Proved)

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More formulas :

Newton's first equation of motion :

• v = u + at

Newton's second equation of motion :

• s = ut + ½ at²

Newton's third equation of motion :

• v² = u² + 2as

Answer 2

Answer:

The first two equations of motion are:

$\sf{v=u+at}$

$\sf{s=ut+\dfrac{1}{2} at^2}$

1st equation, $\sf{v=u+at}$ is available.

$\sf{v^2=(u+at)^2}$

$\sf{v^2=u^2+2aut+a^2t^2}$

$\sf{v^2=u^2+a(2ut+at^2)}$

2nd equation, $\sf{2s=2ut+at^2}$ is derived after multiplying sides by 2.

$\sf{v^2=u^2+a(2s)}$

$\boxed{\sf{\therefore v^2=u^2+2as}}$

Derivation of Equations:

1st eq: Let's derive $\sf{v=u+at}$.

By use of $\sf{a=\dfrac{v-u}{t} }$  [The Definition of Acceleration]

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

→ $\sf{at=v-u}$

Flip sides!

→ $\sf{v-u=at}$

$\boxed{\sf{\therefore v=u+at}}$

2nd eq: The equation of motion $\sf{s=ut+\dfrac{1}{2}at^2 }$.

By use of the Velocity-Time Graph

In an initial acceleration, the graph will form a line.

• v is the final velocity

• u is the initial velocity

• t is time

The area under the graph consists of a right triangle and a rectangle.

And, the area under the graph is the displacement s.

Right Triangle: $\sf{\dfrac{1}{2} \times t\times (v-u)}$

Rectangle: $\sf{ut}$

The first equation, $\sf{v-u=at}$ into the triangle area.

→ $\sf{\dfrac{1}{2} at^2}$ is the area of a right triangle.

→ Therefore, the displacement is $\sf{ut+\dfrac{1}{2} at^2}$.

$\boxed{\sf{\therefore s=ut+\dfrac{1}{2}at^2}}$

Explanation:

See the attachment :