Given the v-t diagram for the motion of a train as it moves from station A to station B

To find:

1. Draw a-t graph
2. find the average speed of the train

3. the distance between the stations

    

From the given graph we can observe that

0s to 30s linear variation of velocity with positive acceleration

30s to 90s constant velocity of 12m/s

90s to 120s linear variation with negative acceleration

 (v=\dfrac{30}{12}t, 0s<t<30s)</p>

(v=12m/s,30s<t<90s)</p>

(v=48-\dfrac{12}{30}x, 90s<t<120s)</p>

We know that $a=\dfrac{dv}{dt}=$slope of v-t graph

so

(a=\dfrac{30}{12},0s<t<30s)</p>

(a=0m/s^2,30s<t<90s)</p>

(a=-\dfrac{12}{30}m/s^2,90s<t<120s)</p>

Plotting the graph

Distance between stations(uniform velocity)$=s=vt=12(60)=720m$

Average speed=$v_a=\dfrac{total\ distance}{total\ time}$

Total distance covered=area under v-t curve=$\dfrac{1}{2}30(12)+12(60)+\dfrac{1}{2}30(12)=180+720+180=1080m$

Total time=T=120s

So average speed=$\dfrac{1080}{120}=9m/s$

So distance between stations A and B is $720m$

Average speed$ =9m/s$