Question

prove that
2aS=Vf2 – Vi2

Answer 1

Answer:

EQUATIONS OF MOTION EQUATIONS OF MOTION

FIRST EQUATION OF MOTION

Vf = Vi + at

Consider a body initial moving with velocity "Vi". After certain interval of time "t", its velocity becomes "Vf". Now

Change in velocity = Vf - Vi

OR

DV =Vf – Vi

Due to change in velocity, an acceleration "a" is produced in the body. Acceleration is given by

a = DV/t

Putting the value of "DV"

a = (Vf – Vi)/t

at = Vf – Vi

at + Vi =Vf

OR

SECOND EQUATION OF MOTION

OR

S = Vit + 1/2at2

Consider a car moving on a straight road with an initial velocity equal to ‘Vi’. After an interval of time ‘t’ its velocity becomes ‘Vf’. Now first we will determine the average velocity of body.

Average velocity = (Initial velocity + final velocity)/2

OR

Vav = (Vi + Vf)/2

but Vf = Vi + at

Putting the value of Vf

Vav = (Vi + Vi + at)/2

Vav = (2Vi + at)/2

Vav = 2Vi/2 + at/2

Vav = Vi + at/2

Vav = Vi + 1/2at.......................................(i)

we know that

S = Vav x t

Putting the value of ‘Vav’

S = [Vi + 1/2at] t

THIRD EQUATION OF MOTION

OR

2aS = Vf2 – Vi2

Initial velocity, final velocity, acceleration, and distance are related in third equation of motion.

Consider a body moving initially with velocity ‘Vi’. After certain interval of time its velocity becomes ‘Vf’. Due to change in velocity, acceleration ‘a’ is produced in the body. Let the body travels a distance of ‘s’ meters.

According to first equation of motion:

Vf = Vi + at

OR

Vf – Vi = at

OR

(Vf – Vi)/a = t....................(i)

Average velocity of body is given by:

Vav = (Initial velocity + Final velocity)/2

Vav = (Vi + Vf)/2.................. (ii)

we know that :

S = Vav x t.................. (ii)

Putting the value of Vav and t from equation (i) and (ii) in equation (iii)

S = { (Vf + Vi)/2} { (Vf – Vi)/a}

2aS = (Vf + Vi)(Vf – Vi)

According to [ (a+b)(a-b)=a2-b2]

Answer 2

Answer:

2aS = $v_f^2 - v_i^2$ proved.

Explanation:

Let final velocity be $v_f$ and let  initial velocity be $v_i$ .

As we know that the formula of  the third law of motion.

$S = ut +\frac{1}{2} at^2$ and v = u + at .

If final velocity = $v_f$ and initial velocity =$v_i$ than,

S = $v_i t + \frac{1}{2} at^2$   ...........(i)

and $v_f = v_i + at$  ..........(ii)

From the equation of  $v_f = v_i + at$ we have,

t = $\frac{v_f - v_i}{a}$.

Now, put the value of 't' in equation (i) we get,

S = $v_i(\frac{v_f- v_i}{a} ) +\frac{1}{2} a(\frac{v_f -v_i}{a} )^2$

⇒S = [$\frac{2(v_iv_f- v^2_i) +(v_f -v_i)^2 }{2a}$]

⇒2aS = $2v_iv_f - 2v_i^2 + (v_f^2 + v_i^2-2v_fv_i)$

⇒2aS = $-2v_i^2 +v_f^2 + v_i^2$

⇒2aS = $v_f^2 - v_i^2$

Hence, here we proved that 2aS = $v_f^2 - v_i^2$ .

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