Question

an object travels along a straight line. the function $s(t) = 5\tan t$ gives the objects position, in miles, at time $t$. the time $t$ is given in hours, with $0 < t < \frac{\pi}{2}$.
write a function that gives the objects acceleration $a(t)$ in miles per hour per hour.
$a(t) = $

Explanation:

Step1: Find first derivative (velocity)

Velocity \(v(t)\) is the derivative of position \(s(t)\).
The derivative of \(\tan t\) is \(\sec^2 t\), so:
\(v(t) = s'(t) = 5\sec^2 t\)

Step2: Find second derivative (acceleration)

Acceleration \(a(t)\) is the derivative of velocity \(v(t)\). Use the chain rule: let \(u = \sec t\), so \(v(t)=5u^2\).
First, \(\frac{d}{dt}(u^2)=2u \cdot u'\), and \(\frac{d}{dt}(\sec t)=\sec t \tan t\).
So:
\(a(t) = v'(t) = 5 \cdot 2\sec t \cdot \sec t \tan t = 10\sec^2 t \tan t\)

Answer:

\(10\sec^2 t \tan t\)