Relative Ages in a Riverbank and a Seismic Distance
Along a riverbank, students record four horizontal sedimentary layers from top to bottom: A, B, C, D. Layer A contains marine shells; B contains rounded pebbles; C contains ammonite fossils; D contains granite fragments. A vertical igneous intrusion cuts through B, C, and D but ends beneath A. A fault cuts layer A and the intrusion, but its displacement dies out above B, so B shows no offset. A thin layer of loose river gravel lies on top of A. Separately, a seismic station records an earthquake: the P wave arrives at 10:00:05 and the S wave at 10:00:12. Assume P-wave speed 6.0 km/s and S-wave speed 3.5 km/s, and that the waves traveled the same straight path.
1. Place the following in order from oldest to youngest: layers A–D, the igneous intrusion, the fault, and the river gravel. Justify each step using superposition or cross-cutting relationships.
Read the worked answer
Order: D → C → B → intrusion → A → fault → river gravel.
- D is lowest, so by superposition it is oldest, followed by C, then B, then A above them.
- The intrusion cuts B, C, and D, so it is younger than those layers. It ends beneath A, so A was deposited after the intrusion.
- The fault cuts A and the intrusion, so it is younger than both. It does not offset B because its displacement dies out above B.
- The river gravel lies on top of A and is loose, so it is the youngest feature.
2. Calculate the distance from the station to the earthquake and the origin time. Show your equation and units.
Read the worked answer
The S–P time difference is 12 − 5 = 7.0 s.
Let d be the distance. Then:
d/3.5 − d/6.0 = 7.0
Using a common denominator: (6d − 3.5d)/21 = 2.5d/21 = 7.0, so d = 7.0 × 21 / 2.5 = 58.8 km.
The P wave traveled 58.8 km at 6.0 km/s, taking 58.8 / 6.0 = 9.8 s. Therefore the origin time was 10:00:05 − 9.8 s ≈ 09:59:55.2, or about 09:59:55.