alright , the first thing u need to know is that the distance travelled is area under the speed-time graph.
c= distance travelled by cheeta in 17 secs = area under its graph upto 17 sec=
=(1/2)*2*29 + 29*15 = 29(16)=464
similarly,
a= distance by antelope= (1/2)*2*25 + 25*14 = 25*15= 375
So, net distance after 17 sec = 100 + a - c =100+375-464 = 11 m
so thats option B
