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.
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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