Scientists determine that sedimentary rock has been depositing at a location at a rate of 5 cm/a.?

a) Determine the age of a strata of sedimentary rock that is 10 km thick.

b) How thick would a one million year old strata be if sediment is deposited at this rate?

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  • 5 years ago
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    a) Determine the age of a strata of sedimentary rock that is 10 km thick.

    Assuming that 5 cm/a means 5 centimeters per annum or 5 centimeters

    per year, then...

    This is a standard "rate" problem where...

    d = depth (or thickness) of a stratum (singular) = 10 km = 10,000 m = 1,000,000 cm

    r = rate of deposition = 5 cm/yr

    t = time required (age) for the stratum to be deposited = to be determined

    r = d/t

    rt = d

    t = d/r

    t = 1,000,000 cm / 5 cm/yr

    t = 200,000 years ANSWER

    b) How thick would a one million year old strata be if sediment is deposited

    at this rate?

    d = depth (or thickness) of a stratum (singular) = to be determined

    r = rate of deposition = 5 cm/yr

    t = time required (age) for the stratum to be deposited = 1,000,000 years

    r = d/t

    rt = d

    d = rt

    d = (5 cm/yr)(1,000,000 yr)

    d = 5,000,000 cm = 50,000 m = 50 km ANSWER

  • Andrew
    Lv 5
    5 years ago

    Possibly a flawed question that confuses age and time intervals...

    a) the age wound vary linearly with depth between 200,000 years (10 000 m/0.05 m/yr) at the bottom of the stratum, and 0 at the top, assuming it is still being deposited. Note that you specify 'age' not 'total time for deposition', which I think perhaps is what you meant to ask.

    b) the thickness would effectively be zero. This is again because the age of the stratum must vary with depth given your 0.05 m/yr deposition rate. As you specify a 1 million year 'age' NOT a 1 million year deposition period, you are actually asking how much material was deposited instantaneously 1 million years ago (which mathematically is 0 m), not how much was deposited over a one million year period, which I think is perhaps what you really meant.

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