Given, vAB=7m∕s,(v1)BC=14m∕s,(v2)BC=21m∕s The distance between A and B is, dAB=vABt dAB=7t The distance between B and C is, dBC=(vBC)argt′ dBC=[
(vBC)1+(vBC)2
2
]t′ dBC=[
14+21
2
]t′ dBC=
35
2
t′ According to question, dAB=dBC 7t=
35
2
t′ t′=
2
5
t The total time taken from A to C is, T=t+t′ T=t+
2t
5
T=
7t
5
The distance AC is, dAC=vt dAC=[
vAB+(vBC)1+(vBC)2
3
]t dAC=[
7+14+21
3
]t dAC=14t The average velocity during the whole journey is , vavg=