The velocity-time graph of a body moving in a straight line is shown in the figure. Find the displacement and distance travelled by the body in 10 seconds.
Show Hint
The displacement is the net area under the velocity-time graph, while the distance is the sum of all areas treated as positive.
Step 1: Displacement calculation.
The displacement is the area under the velocity-time graph. The graph consists of two parts:
1. A triangle from 0 to 4 seconds with a peak at 20 m/s, giving an area of:
\[
\text{Area of triangle} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \, \text{s} \times 20 \, \text{m/s} = 40 \, \text{m}
\]
2. A trapezoid from 4 to 10 seconds, with the heights of 0 and -20 m/s at the endpoints and a base of 6 seconds, giving an area of:
\[
\text{Area of trapezoid} = \frac{1}{2} \times (20 + 0) \times 6 = 60 \, \text{m}
\]
Thus, the total displacement is:
\[
\text{Displacement} = 40 \, \text{m} + 60 \, \text{m} = 50 \, \text{m}
\]
Step 2: Distance travelled.
The distance travelled is the total area under the graph, treating negative values as positive. The total area is the sum of the absolute values of both areas:
\[
\text{Distance travelled} = 40 \, \text{m} + 60 \, \text{m} = 90 \, \text{m}
\]
Final Answer:
\[
\boxed{50 \, \text{m}, 90 \, \text{m}}
\]
