Question

A lake near the Arctic Circle is covered by a 2-meter-thick sheet of ice during the cold winter months. When spring arrives, the warm air gradually melts the ice, causing its thickness to decrease at a constant rate. After 3 weeks, the sheet is only 1.25 meters thick. Let S(t) denote the ice sheet's thickness S (measured in meters) as a function of time t (measured in weeks). Write the function's formula.

Answer 1

Answer: The required function formula is,

S(t) = 2 - 0.25 t

Step-by-step explanation:

Here, the initial thickness of the ice sheet = 2 meters,

After 3 weeks, the thickness of ice sheet = 1.25 meters

Total changes in the thickness in 3 weeks = 2 - 1.25 = 0.75 meters,

⇒ Total changes in the thickness in 1 weeks = 0.75/3 = 0.25 meters,

Since, the ice is melting with the constant rate.

⇒ The rate of ice decreasing = 0.25 meters per week.

⇒ Total changes in t weeks = 0.25 t meters

The new thickness of ice after t weeks = Initial thickness - Total changes in t weeks.

⇒ S(t) = 2 - 0.25 t

Which is the required function's formula.

Answer 2

A lake near the Arctic Circle is covered by a 2-meter-thick sheet of ice during the cold winter months.

When spring arrives, the warm air gradually melts the ice, causing its thickness to decrease at a constant rate.

S(t) denote the ice sheet's thickness S ( measured in meters) as a function of time (measured in weeks).

Therefore the equation formed will be linear.

The equation will be of the form y = mx + b

Here S(t) = mt + b

Here m is the slope which is the rate at which ice is melting.

Putting t = 0

S(t) = 2

Putting t = 3,

S(t) = 1.25

Therefore, m*0 + b = 2 or, b = 2

and 3m + b = 1.25

or, 3m = 1.25 - 2 = -0.75

or, t = -0.25

Hence, function's formula = S(t) = -0.25*t + 2

i.e. S(t) = 2 - 0.25*t