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Is it true that the third equation of motion is $v^{2} - u^{2} = 2 as$ ?

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True

False

answer is A.

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Detailed Solution

The above statement is true.
The third equation of motion is found from the area under the velocity time graph. The third equation of motion connects the final velocity

(v)

, the initial velocity

(u)

, the acceleration of a body

(a)

, and the displacement of the body

(s)

. The third equation of motion is also called the equation for position velocity relation.
Question Image

We can deduce from the graph that
The Area of trapezium

OABC

gives the total distance travelled,

s .

Hence,
Height

(s) = 12

(Sum of Parallel Sides)

\frac{1}{2} (OA + CB) \times OC = s

Because

OA = u, CB = v

, and

OC = t

The above equation is transformed into

\frac{1}{2} \times (u + v) \times t = s

Now, because

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

The preceding equation can be expressed as follows:

\frac{1}{2} \times \frac{((u + v) (v - u))}{a} = s

We get by rearranging the equation

\frac{1}{2} \frac{(v + u) (v - u)}{a} = s

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

Solving the above equation yields the third equation of motion:

v^{2} = u^{2} + 2 aS

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