If a layer was deposited but does not appear in the rock record, what has occured
2 answers:
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Refer to the diagram shown below.
d = distance (miles) from Alphaville to Betaville.
v = speed (mph) of the plane with no wind.
With no wind:
The time taken to travel a distance of 2d is 4 hrs, 48 min = 4.8 hrs.
Therefore
2d/v = 4.8
v = 2d/4.8 = 0.4167d mph (1)
With the wind:
The velocity from Alphaville to Betaville is (v + 100) mph.
The time of travel is
t₁ = d/(v+100) h
The velocity from Betaville to Alphaville is (v - 100) mph.
The time of travel is
t₂ = d/)v-100) h
Because the return trip takes 5 hours, therefore
t₁ + t₂ = 5
(2)
From (1), obtain
2(0.4167)d² = 5[(0.4167d)² - 10⁴]
0.8334d² = 0.8682d² - 5 x 10⁴
0.0348d² = 5 x 10⁴
d = 1198.7 mi
Answer: 1199 miles (nearest integer)
d = distance (miles) from Alphaville to Betaville.
v = speed (mph) of the plane with no wind.
With no wind:
The time taken to travel a distance of 2d is 4 hrs, 48 min = 4.8 hrs.
Therefore
2d/v = 4.8
v = 2d/4.8 = 0.4167d mph (1)
With the wind:
The velocity from Alphaville to Betaville is (v + 100) mph.
The time of travel is
t₁ = d/(v+100) h
The velocity from Betaville to Alphaville is (v - 100) mph.
The time of travel is
t₂ = d/)v-100) h
Because the return trip takes 5 hours, therefore
t₁ + t₂ = 5
From (1), obtain
2(0.4167)d² = 5[(0.4167d)² - 10⁴]
0.8334d² = 0.8682d² - 5 x 10⁴
0.0348d² = 5 x 10⁴
d = 1198.7 mi
Answer: 1199 miles (nearest integer)