Question

From the velocity-time plot shown in figure, find the distance traveled by the particle during the first $40$ seconds. Also find the average velocity during this period.

Answer 1

$\text{Step 1: Distance calculation}$

The area under the velocity-time graph is displacement.

Here upward area is positive and downward area is negative. But for distance we will take both area as positive because distance is always positive.

Distance in first $40sec$ is $\Delta OAB+\Delta BCD$

$D=\frac{1}{2}\times 20\times 5+|-\frac{1}{2}\times 20\times 5|$ $m$ (Distance is always positive)

$=100m$

$\text{Step 2: Average velocity calculation}$

Average velocity $=\frac{TotalDisplacement}{time}$

Displacement $=50m+(-50m)=0$

$\therefore$ Average velocity $=0$

Hence distance travelled during first $40s$ is $100m$ and average velocity is $0$.