Question

1. if s(t) denotes the position of an object in feet at time t seconds, s(t) is the acceleration of the object in ft/s² at time t seconds. if s(t) = t⁴/³ − 7t + 4 ft, find the acceleration at time t = 8 s.

Explanation:

Step1: Find the first derivative \( s'(t) \) (velocity)

The position function is \( s(t) = t^{4/3} - 7t + 4 \). Using the power rule \( \frac{d}{dt}t^n = nt^{n - 1} \), we have:
\( s'(t)=\frac{4}{3}t^{\frac{4}{3}-1}-7=\frac{4}{3}t^{\frac{1}{3}} - 7 \)

Step2: Find the second derivative \( s''(t) \) (acceleration)

Differentiate \( s'(t) \) with respect to \( t \):
\( s''(t)=\frac{4}{3}\times\frac{1}{3}t^{\frac{1}{3}-1}-0=\frac{4}{9}t^{-\frac{2}{3}}=\frac{4}{9t^{\frac{2}{3}}} \)

Step3: Evaluate \( s''(t) \) at \( t = 8 \)

Substitute \( t = 8 \) into \( s''(t) \):
\( s''(8)=\frac{4}{9\times8^{\frac{2}{3}}} \)
First, calculate \( 8^{\frac{2}{3}}=(8^{\frac{1}{3}})^2 = 2^2 = 4 \)
Then, \( s''(8)=\frac{4}{9\times4}=\frac{1}{9} \)

Answer:

The acceleration at \( t = 8 \) seconds is \(\frac{1}{9}\) \( \text{ft/s}^2 \)