Answer:
[tex]1300 m[/tex]
Explanation:
As the path is straight, so the speed is equivalent to velocity. Now. assuming that the acceleration and deceleration of the train are constant. So, change of velocity with respect to time for acceleration as well as deceleration is constant. Hence, the slope of the speed-time graph is constant for the time of acceleration as well as deceleration. The speed for the time from [tex]20 s[/tex] to [tex]70 s[/tex] is constant, so slope for this interval of time is zero. The speed-time graph is shown in the figure.
The total distance covered by the train during the entire journey is the area of the speed-time graph.
Area[tex]=\frac{1}{2}(20\times 20)+ 50\times 20+\frac{1}{2}(20\times 10)[/tex]
[tex]=200+1000+100[/tex]
[tex]=1300[/tex]
As velocity is in [tex]m/s[/tex] and time is in [tex]s[/tex] so the unit of area is [tex]m[/tex]
Hence, the total distance is [tex]1300m[/tex].
