The area of rectangle OADC is the base multiplied by height: OC × OA = t × u = ut. For the triangle ABD, the base is AD = OC = t, and the height is BD = BC - CD = v - u.

Substituting BD = at and AD = t into the area of triangle ABD, we get (1)/(2) × t × at = (1)/(2)at². Adding this to the area of rectangle OADC, we obtain the second equation of motion: s = ut + (1)/(2)at².

Substituting BD = at and AD = t into the area of triangle ABD, we get (1)/(2) × t × at = (1)/(2)at². Adding this to the area of rectangle OADC, we obtain the second equation of motion: s = ut + (1)/(2)at².

Physics mein UT plus half year ko graphics method se humko samjhaen

Answer: By graphical method, the displacement is equal to the area under the velocity-time graph: $s = \text{Area}\text{(OADC)} + \text{Area}(\triangle ABD) = ut + \frac{1}{2}at^2$ .

Step-by-step solution

Step 1: Set up the Velocity-Time graph

Consider an object moving along a straight line with uniform acceleration $a$ . On a velocity-time graph, the initial velocity at time $t = 0$ is represented by the point $A$ such that $OA = u$ . At time $t = OC$ , the final velocity reaches $v = BC$ .

Step 2: Relate displacement to total area under the graph

The distance or displacement $s$ traveled by the object in time $t$ is equal to the area bounded under the line $AB$ and the time axis. We split the trapezium OABC into two simple shapes: rectangle OADC and right-angled triangle $ABD$ .

Step 3: Calculate the areas of the rectangle and triangle

The area of rectangle OADC is the base multiplied by height: $OC \times OA = t \times u = ut$ . For the triangle $ABD$ , the base is $AD = OC = t$ , and the height is $BD = BC - CD = v - u$ .

Step 4: Substitute acceleration formula into the triangle area

By definition, uniform acceleration $a$ is the rate of change of velocity, which is the slope of the graph: $a = \frac{BD}{AD} = \frac{BD}{t}$ . Therefore, the vertical change $BD$ is equal to $at$ .

Step 5: Combine the areas to derive the final equation

Substituting $BD = at$ and $AD = t$ into the area of triangle $ABD$ , we get $\frac{1}{2} \times t \times at = \frac{1}{2}at^2$ . Adding this to the area of rectangle OADC, we obtain the second equation of motion: $s = ut + \frac{1}{2}at^2$ .

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