A ball is thrown into a lake, creating a circular ripple that travels outward at a speed of 5 cm per second. Express the area, A. of the circle in terms of the number of seconds, t, that have passed since the ball hits the lake. A = 25t A = 2576 A = Ar? A = 5t? None of the above

The area, A. of the circle in terms of the number of seconds, t, that have passed since the ball hits the lake is A(t)= 25t²π

The formula for calculating the area of the circular ripple as a function of time is expressed as:

A(t) = πr(t)²

Given that the ripple travels outward at a speed of 5 cm per second, hence;

r(t) = 5tcm/s

Substitute the radius into the formula to have:

A(t) = π(5t)²

A(t)= 25t²π

Hence the area, A. of the circle in terms of the number of seconds, t, that have passed since the ball hits the lake is A(t)= 25t²π

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The area of the circular ripple after t second is

A =25π t²

where A is in cm².

Step-by-step explanation:

Given that,

Circular ripple that travels outward at a speed of 5 cm per second.

It means that, the radius of the circular ripple increases at a seed 5 cm/ s.

Therefore the radius of the circular ripple after t seconds is

r(t)= 5t

where r(t) is in cm.

We know ,

The area of a circular object is = π r²

The area of the circular ripple after t second is

A= π(5t)²

 =25πt²

where A is in cm².

300pi

step-by-step explanation: