Question

Graphically, derive the first equation of motion v = u + at where ‘u’,’v’,’a’ and ‘t’ has their usual meaning.

Answer 1

Graphically Derivation of First Equation of Motion :

Consider the velocity time graph of a body shown in attachment. The body has an initial Velocity u at point A and then velocity changes from A to B at a uniform rate in time t. There's uniform acceleration a from A to B,and after time t the final Velocity is v which is equal to BC. The time t is represented by OC. Through construction we drew perpendicular CB from point C,and AD parallel to OC. BE is perpendicular to the point B to OE.

Now,

• Initial Velocity (u) = OA... i)

• Final Velocity (v) = BC....ii)

From the graph BC = BD + DC

$\therefore$ v = BD + DC ....iii)

Again, DC = OA

So,

v = BD + OA

From equation i), OA = u

So,

v = BD + u ....iv)

We need to find the value of BD now. We know that Slope of velocity time graph is equal to the acceleration,a.

Thus,

Acceleration = a = slope of line AB

→ a = BD/AD

AD = OC = t,so putting t in the place of AD

→ a = BD/t

→ BD = at

Putting the value of BD in equation iv)

→ v = BD + u

→ v = at + u

→ v = u + at

And this is first equation of motion. It has been derived here by the graphical method.

Answer 2

We need to graphically derive the first equation of motion having the relationship between final velocity ,initial velocity ,acceleration and time.

Let's consider a case :

An object has initial velocity u , experiences acceleration a which leads to its increase in velocity. Let its final velocity be v and the time taken for this transformation of velocity be t

Assumption:

The most important assumption in this derivation is that :

The acceleration "a" constant in value.

Derivation :

Let's draw a graph with acceleration on Y axis and time on X axis :

We know that area under the acceleration vs time graph gives us the change in velocity.

So , we can say :

$\sf{Area \: of \: Rectangle = \Delta v }$

$= > \sf{a \times t = \Delta v }$

$= > \sf{(a \times t) = v - u }$

$= > \sf{v = u + at }$

[ Hence proved ]