Question

derive the second equation of motion by graphical method​

Answer 1

To Derive 2nd equation of motion [Graphically]

★ s = ut + ½ at²

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.

Here,

• Initial Velocity (u) = OA

• Final Velocity (v) = BC

• Time (t) = OC

Distance Covered = Area of the figure OABC

Area of OABC = Area of OADC + Area of ABD

→ Distance (s) = Area of OADC + Area of ABD

→ s = (OA × OC) + (½ × AD × BD)

→ s = (u × t) + (½ × t × at)

→ s = ut + ½ at²

$\therefore$ The 2nd equation of motion is derived here graphically.

Answer 2

Answer:

• Suppose an object travels in a straight line with uniform acceleration $a$.
• Let $v$ be the object's final velocity at the time $t$, and let $u$ be the object's initial velocity at the time $t=0$.
• Let $s$  be the distance covered by the object over time $t$.
• The object's motion is a straight line $PQ$ in the velocity-time graph as shown in the figure, where,

       $OP=RS=u$

       $OW=SQ=v$

       $OS=PR=t$

We know that the distance traveled by a uniformly accelerated object at a given time interval is represented by the area under the velocity-time graph for that time period.

∴ Distance (displacement) traveled by the object in time $t$ is :

$S = Area of Trapezium OSQP$

$S= Area of Rectangle OSRP + Area of Triangle PRQ$

$S= OS$ × $OP+\frac{1}{2}$ × $PR$ × $PQ$

Now putting the values, we get,

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

Now from the first equation of motion, $v-u=at$

$S=ut+\frac{1}{2}$ × $t$ × $at$

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

This is the second equation for uniformly accelerated motion.

#SPJ2