From the given graph find the distance and displacement."

Distance=Area of triangle 1+Area of rectangle 2+Area of triangle 3+Area of the triangle 4

$=\frac{1}{2}\times2\times10+2\times10+\frac{1}{2}\times1\times10+\frac{1}{2}\times1\times10$

$=10+20+5+5$

$=40$

Therefore, distance$=40\ m$

For displacement:

In $\boxed{1}$, change of velocity $\Delta v=v_2-v_1=10-0=10\ m$

Change in time $\Delta t=t_2-t_1=2-0=2\ s$

Therefore, displacement $d_1=\Delta v\times\Delta t=10\times2=20\ m$

In $\boxed{2}$, there is a uniform speed,

So, change in speed $\Delta v=v_2-v_1=10-10=0$

Change in time $\Delta t=t_2-t_1=4-2=2$

So, the displacement $d_2=\Delta v\times\Delta t=0\times2=0$

In $\boxed{3}$,

Change in velocity $\Delta v=v_2-v_1=0-10=-10\ m/s$

Change in time $\Delta t=t_2-t_1=5-4=1\ s$

So, the displacement $d_3=\Delta v\times \delta t=-10\times1=-10\ m$

In $\boxed{4}$,

Change in velocity $\Delta v=v_2-v_1=-10-0=-10\ m/s$

Change in time $\Delta t=t_2-t_1=6-5=1\ s$

So the displacement $d_4=\Delta v\times\Delta t$

$=-10\times1=-10\ m$

Total displacement in the graph$=d_1+d_2+d_3+d_4$

$=20+0-10-10$

$=0$

Thus, from the given graph, we find total distance$=40\ m$ and the displacement$=0$.

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Updated on: 10-Oct-2022

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