Question

S=ut+1/2at2 prove by graphical method.S

Answer 1

Given that,

Second equation of motion.

Suppose a body has initial velocity u at point A and this velocity changes to v at point B in t secs.

So, The final velocity will be v.

We need to calculate the distance covered by the object

Using diagram

$s=\text{Area of OABC}$

$s=\text{Area of OADC}+\text{Area of ABD}$

$s=OA\times AD+\dfrac{1}{2}\times AD\times BD$

Put the value into the formula

$s=u\times t+\dfrac{1}{2}\times(v-u)$

We know that,

$v=u+at$

$a=\dfrac{v-u}{t}$

Put the value of (v-u) from first equation of motion

$s=ut+\dfrac{1}{2}\times at^2$

Hence, This is second equation of motion.

Learn more :

Topic : equation of motion

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Answer 2

Concept: There are three equations for the motion of those bodies which travel with uniform acceleration. These equations give relationship between initial velocity, final velocity, time taken, acceleration and distance travelled by the bodies.

To find: $S=ut+\frac{1}{2} at^{2}$

Given: The second equation of motion is $S=ut+1/2at^{2}$ . It gives the distance travelled by a body in time t.

Solution: Suppose a body has an initial velocity 'u' and a uniform acceleration 'a' for time 't' so that its final velocity becomes 'v'. Let the distance travelled by the body in this time be 'S'. The distance travelled by the body is given by the area of the space between the velocity-time graph AB and the time axis OC, which is equal to the area of the figure OABC. Thus,

Distance travelled = Area of figure OABC

                   = Area of rectangle OADC + Area of triangle ABD

First, we will find out the area of the rectangle OADC and the area of triangle ABD

$Area of rectangle = OA * OC\\ = u*t\\ = ut$

$Area of triangle ABD = \frac{1}{2} * Area of rectangle AEBD\\ = \frac{1}{2} *AD * BD\\ =\frac{1}{2} *t * at (because AD=t and BD= at)\\ = \frac{1}{2} at^{2}$

So, Distance travelled S= Area of rectangle OADC + Area of triangle ABD

or, $S = ut + \frac{1}{2} at^{2}$

This is the second equation of motion. It has been derived here by the graphical method.

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