AP EAMCET Engineering 20 Apr 2019 Shift 1 Paper

Let vp is the velocity of P wave, vs is the velocity of S wave and d is the distance of epicenter from the point of observation. Then,
d=vptp=vsts
tp=

vs

vp

ts
tp=

4.5

8

ts
It is given that the first P wave arrives 3.5 min earlier than the first S -wave.
ts−tp‌‌=3.5×60
ts−

4.5

8

ts‌‌=210
ts‌‌=

210×8

3.5

‌‌=480s
The distance d comes out to be,
d‌‌=vsts
‌‌=4.5×480=2160km