Question

Using Velocity- time graph derive the three equations of motion.
 v = u + at
 s = ut + ½ at^2
 v^2 = u2 + 2as

Answer 1

Step-by-step explanation:

i think it may helpful to uh ☺️

Answer 2

Step-by-step explanation:

Derivation of First Equation of Motion by Graphical Method

The first equation of motion can be derived using a velocity-time graph for a moving object with an initial velocity of u, final velocity v, and acceleration a.

See image

the above graph,

• The velocity of the body changes from A to B in time t at a uniform rate.

• BC is the final velocity and OC is the total time t.

• A perpendicular is drawn from B to OC, a parallel line is drawn from A to D, and another perpendicular is drawn from B to OE (represented by dotted lines).

Following details are obtained from the graph above:

The initial velocity of the body, u = OA

The final velocity of the body, v = BC

From the graph, we know that

BC = BD + DC

Therefore, v = BD + DC

v = BD + OA (since DC = OA)

Finally,

v = BD + u (since OA = u) (Equation 1)

Now, since the slope of a velocity-time graph is equal to acceleration a,

So,

a = slope of line AB

a = BD/AD

Since AD = AC = t, the above equation becomes:

BD = at (Equation 2)

Now, combining Equation 1 & 2, the following is obtained:

${ {\boxed{ \rm v = u + at}}}$

Derivation of Second Equation of Motion by Graphical Method

See 2nd image

From the graph above, we can say that

Distance travelled (s) = Area of figure OABC = Area of rectangle OADC + Area of triangle ABD

$\rm s = \frac{1}{2} AB×BD+(OA×OC)$

Since BD = EA, the above equation becomes

$\rm s = ( \frac{1}{2} AB × EA) + (u×t)$

As EA = at, the equation becomes

$\rm s = \frac{1}{2} \times at \times t + ut$

On further simplification, the equation becomes

$\boxed{ \rm s = ut + \frac{1}{2} a {t}^{2} }$

Derivation of Third Equation of Motion by Graphical Method

See 3rd image

From the graph, we can say that

The total distance travelled, s is given by the Area of trapezium OABC.

Hence,

$\rm S = \frac{1}{2} (Sum \: of \: Parallel \: Sides) × Height$

$\rm S=(OA+CB)×OC$

Since, OA = u, CB = v, and OC = t

The above equation becomes

$\rm S= \frac{1}{2} (u+v) × t$

Now,

$\rm since \: t = \frac{(v – u)}{a}$

The above equation can be written as:

$\rm S= \frac{1}{2} \frac{((u+v) × (v-u))}{a}$

Rearranging the equation, we get

$\rm S= \frac{(v+u) × (v-u)}{2a}$

$\rm S = \frac{ (v ^{2} -u^{2})}{2a}$

Third equation of motion is obtained by solving the above equation:

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