The correct option is
B The particle crosses its initial position at
t=2 s.
Area under v−t graph=Displacement Displacement in time interval
0 to 1 s =Area(ΔOAB) =12×1×10=5 m ...(1) ![](https://search-static.byjusweb.com/question-images/byjus/ckeditor_assets/pictures/944915/original_Qn2_s.png)
Displacement in time interval
1 to 2 s =Area(ΔBCD) =−12×1×10=−5 m ...(2) Now, net displacement (sum of total area under curve) from
t=0 to
2 s is,
S=+5+(−5)=0 ∴ Particle will cross its initial position at
t=2 s, since net displacement of particle is zero at the end of
t=2 s.
Total distance covered in
0≤t≤2 s is not zero. So, average speed is also not zero.
And initial speed of particle is not zero. (from the graph)
Hence, only option (b) is correct.