**Derive the third equation of motion for uniformly accelerated rectilinear motion.**

**OR**

**Which equation of motion establishes the relationship between S, a, V _{i }and V_{f}?**

**OR**

**Derive the equation of motion which is independent of t.**

**OR**

**Derive the third equation of motion?**

**OR**

**Prove that $2aS = v_f^2 - v_i^2$**

Difficulty: Hard

**Suppose a body is moving with initial velocity v _{i} and after a certain time t its velocity becomes v_{f} then the total distance S covered in time t is given by**

S = ${vav}\times t$

S = $\frac{vf \: + \: vi}{2}\times t$ ……..(1)

From the first equation of motion find the value of t.

vf = vi +at Or t = $S=\frac{vf \: - \: vi}{a}$

Putting the value of V_{f} in equation (1).

S = $\frac{vf \: + \: vi}{2}\times\frac{vf \: - \: vi}{a}$

2as = $(vf+vi)\times (vf - vi)$ by using formula 2as = $(a+b)(a-b)=a^{2}-b^{2}$

2as = $vf^{2}-vi^{2}$

**Second Method (Graphical method)**

**Third equation of motion:**

In the speed-time graph shown in the figure, the total distance S traveled by the body is given by the total area OABD under the graph.

Total area $OABD = S=\left(\frac{OA + BD}{2} \right)\times OD$

Or $2S = (OA + BD) x OD$

Multiply both sides by BC/OD, we get:

$\frac{BC}{OD} =a$

$2S\times\frac{BC}{OD} =\left(OA + BD\right)\times OD\times\frac{BC}{OD}$

$2S\times\frac{BC}{OD} =\left(OA + BD\right)\times BC$ .........(1)

Putting the value in the above equation (1), we get

$2S\times a =\left(Vi + Vf\right)\times \left(Vf - Vi\right)$

$2aS = Vf^{2} + Vi^{2}$

**USEFUL INFORMATION**

- To convert ms
^{-1}to kmh^{-1}

1 ms^{-1} = 0.001km x 3600 = 3.6 kmh^{-1}

Thus, multiply speed in ms^{-1} by 3.6 to get speed in km^{-1} e.g.,

20 ms^{-1}= 20 x 3.6 kmh^{-1}=72 kmh^{-1}

- To convert kmh
^{-1 }to ms^{-1}

^{$1 kmh^{-1}= \frac{1000m}{60x60s} =\frac{10}{36}ms^{-1}$}

Thus, multiply speed in kmh^{-1 }by to get speed in ms^{-1} e.g.

$50 kmh^{-1}= 50 \times \frac{10}{36} ms^{-1}= 13.88 ms^{-1}$

- To convert ms
^{-2}to kmh^{-2}

Multiply acceleration in ms^{-2} by $\frac{3600\times3600}{1000} = 12960$ to get its value in kmh^{-2}

- To convert km
^{-2 }to ms^{-2}

Divide acceleration in kmh^{-2} by 12960 to get its value in ms^{-2}

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