Question

an object travels along a straight line. the function s(t) = √(ln t + 1) gives the object’s position, in miles, at time t > 1 hours. write a function that gives the object’s velocity v(t) in miles per hour.

Explanation:

Step1: Recall the relationship between position and velocity

Velocity is the derivative of the position function with respect to time. So we need to find \( v(t)=\frac{d}{dt}s(t) \), where \( s(t)=\sqrt{\ln t + 1} \).

Step2: Rewrite the position function

Rewrite \( s(t) \) as \( s(t)=(\ln t + 1)^{\frac{1}{2}} \).

Step3: Apply the chain rule

The chain rule states that if we have a function \( y = u^n \), then \( \frac{dy}{dt}=n u^{n - 1}\frac{du}{dt} \). Let \( u=\ln t+1 \) and \( n=\frac{1}{2} \). First, find \( \frac{du}{dt} \): the derivative of \( \ln t \) with respect to \( t \) is \( \frac{1}{t} \) and the derivative of 1 is 0, so \( \frac{du}{dt}=\frac{1}{t} \). Then, find \( \frac{dy}{dt} \): \( n u^{n - 1}=\frac{1}{2}(\ln t + 1)^{-\frac{1}{2}} \). Multiply by \( \frac{du}{dt} \): \( v(t)=\frac{1}{2}(\ln t + 1)^{-\frac{1}{2}}\cdot\frac{1}{t} \).

Step4: Simplify the expression

Simplify \( (\ln t + 1)^{-\frac{1}{2}}=\frac{1}{\sqrt{\ln t + 1}} \), so \( v(t)=\frac{1}{2t\sqrt{\ln t + 1}} \).

Answer:

\( \frac{1}{2t\sqrt{\ln t + 1}} \)