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