The diagram below (Fig. 1.11) illustrates the velocity-time graph of the motion of a body. Calculate the total distance covered by the body.

The parallel sides are the velocities at $t=0 \text{ s}$ and $t=5 \text{ s}$.
The height is the time interval from $t=0 \text{ s}$ to $t=5 \text{ s}$.

$$\text{Area A} = \frac{1}{2} (\text{initial velocity} + \text{final velocity}) \times \text{time interval}$$ $$\text{Area A} = \frac{1}{2} (10 \text{ ms}^{-1} + 20 \text{ ms}^{-1}) \times (5 \text{ s} - 0 \text{ s})$$ $$\text{Area A} = \frac{1}{2} (30 \text{ ms}^{-1}) \times 5 \text{ s}$$ $$\text{Area A} = 15 \text{ ms}^{-1} \times 5 \text{ s} = 75 \text{ m}$$

The length is the constant velocity, and the width is the time interval.

$$\text{Area B} = \text{velocity} \times \text{time interval}$$ $$\text{Area B} = 20 \text{ ms}^{-1} \times (10 \text{ s} - 5 \text{ s})$$ $$\text{Area B} = 20 \text{ ms}^{-1} \times 5 \text{ s} = 100 \text{ m}$$

The parallel sides are the velocities at $t=10 \text{ s}$ and $t=15 \text{ s}$.
The height is the time interval from $t=10 \text{ s}$ to $t=15 \text{ s}$.

$$\text{Area C} = \frac{1}{2} (\text{initial velocity} + \text{final velocity}) \times \text{time interval}$$ $$\text{Area C} = \frac{1}{2} (20 \text{ ms}^{-1} + 0 \text{ ms}^{-1}) \times (15 \text{ s} - 10 \text{ s})$$ $$\text{Area C} = \frac{1}{2} (20 \text{ ms}^{-1}) \times 5 \text{ s}$$ $$\text{Area C} = 10 \text{ ms}^{-1} \times 5 \text{ s} = 50 \text{ m}$$

The total distance is the sum of the areas of sections A, B, and C.

$$\text{Total Distance} = \text{Area A} + \text{Area B} + \text{Area C}$$ $$\text{Total Distance} = 75 \text{ m} + 100 \text{ m} + 50 \text{ m}$$ $$\text{Total Distance} = 225 \text{ m}$$

The total distance covered by the body is $225 \text{ m}$.