Question
Show by using the graphical method that:
$\text{s}=\text{ut}+\frac{1}{2}\text{at}^2$
where the symbols have their usual meanings.
$\text{s}=\text{ut}+\frac{1}{2}\text{at}^2$
where the symbols have their usual meanings.
✓
Answer
Consider the velocity-time graph of a body shown in figure. The body has an initial velocity $u$ at a point $A$ and then its velocity changes at a uniform rate from $A$ to $B$ in time $t.$ In other words, there is a uniform acceleration a from $A$ to $B,$ and after time $t$ its final velocity becomes $v$ which is equal to $BC$ in the graph. The time $t$ is represented by $OC.$
Suppose the body travels a distance $s$ in time $t.$ In the figure, 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 $OABC\ +$ area of triangle $ABD$
Now, we will find out the area of rectangle $OABC$ and area of triangle $ABD.$
$= ut$
$=\Big(\frac{1}{2}\Big)\text{at}^2$
Distance travelled, $s =$ Area of rectangle $OADC\ +$ Area of triangle $ABD$
$\text{s}=\text{ut}+\frac{1}{2}\text{at}^2$
Suppose the body travels a distance $s$ in time $t.$ In the figure, 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 $OABC\ +$ area of triangle $ABD$
Now, we will find out the area of rectangle $OABC$ and area of triangle $ABD.$
- Area of rectangle $OADC = OA × OC$
$= ut$
- Area of triangle ABD $=\Big(\frac{1}{2}\Big)\times\text{Area of rectangle AEBD}$
$=\Big(\frac{1}{2}\Big)\text{at}^2$
Distance travelled, $s =$ Area of rectangle $OADC\ +$ Area of triangle $ABD$
$\text{s}=\text{ut}+\frac{1}{2}\text{at}^2$
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