Question

when an object is dropped, the distance it falls in t seconds, assuming that air resistance is negligible, is given by s(t) = 4.905t² where s(t) is in meters (m). if a stone is dropped from a cliff, assuming that air resistance is negligible, find the velocity and acceleration of the stone after it had been falling for 15 sec. the velocity of the stone is \\( \square \frac{\text{m}}{\text{sec}} \\) (type an integer or decimal rounded to two decimal places as needed.)

Explanation:

Step1: Find the velocity function (derivative of s(t))

The distance function is \( s(t) = 4.905t^2 \). The velocity \( v(t) \) is the first derivative of \( s(t) \). Using the power rule, the derivative of \( t^n \) is \( nt^{n - 1} \), so \( v(t)=\frac{d}{dt}(4.905t^2)=4.905\times2t = 9.81t \).

Step2: Calculate velocity at t = 15 sec

Substitute \( t = 15 \) into the velocity function: \( v(15)=9.81\times15 \).
\( v(15)=147.15 \)

Step3: Find the acceleration function (derivative of v(t))

The acceleration \( a(t) \) is the derivative of \( v(t) \). Since \( v(t)=9.81t \), the derivative is \( a(t)=\frac{d}{dt}(9.81t)=9.81 \). (Acceleration is constant here, so it's 9.81 m/sec² for any t, including t = 15)

Answer:

The velocity of the stone is \( \boldsymbol{147.15} \) \( \frac{\text{m}}{\text{sec}} \), and the acceleration is \( \boldsymbol{9.81} \) \( \frac{\text{m}}{\text{sec}^2} \). (For the velocity part as asked in the box, the answer is 147.15)