Pand s waves from an earthquake travel at different speeds, and this difference in locating the earthquake âepicenterâ (where the disturbance took place). (a) if an earthquake occurs and these two waves arrive at a particular seismic station with the p wave 2.0 min ahead of the s wave, then how far away was the earthquake? assume typical speeds of 8.5 km/s and 5.5 km/s for p and s waves, respectively.

Although the speeds are different, but the two waves will reach the same station so the same distance.

dp = ds

The formula for distance given velocity and time is:

d = v t

Substituting this to the 1st equation:

vp tp = vs ts

Since p wave arrived 2 min (120 s) earlier than s wave, therefore

tp = ts - 120

Then,

vp (ts – 120) = vs ts

8.5 (ts – 120) = 5.5 ts

8.5 ts - 1020 = 5.5 ts

3 ts = 1020

ts = 340 s

 

tp = ts – 120 = 340 – 120 = 220 s

 

Then using either dp or ds, we can calculate how far away the earthquake is:

dp = vp tp

dp = 8.5 * 220

dp = 1870 km = 1.9 * 10^3 km

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I'm having some difficulty seeing the diagram clearly.but the question says there's a net inward force on the object.so we know right away,  from newton's 2nd law of motion, thatthat baby is accelerating.  it doesn't even matter which way thenet force is pointing.  if there's any net force, then either theobject's speed or else its direction must be changing.