Question

the function ( s(t) ) describes the position of a particle moving along a coordinate line, where ( s ) is in feet and ( t ) is in seconds.

( s(t)=\frac{1}{12} t^{2}-ln (t + 1), quad t geq 0 )

(a) find the velocity and acceleration functions.

(b) find the position, velocity, speed, and acceleration at ( t = 1 ).

(c) at what times is the particle stopped? enter as a comma - separated list.

(d) when is the particle speeding up? slowing down? enter using interval notation.

Explanation:

Step1: Find velocity function

The velocity function \(v(t)\) is the derivative of the position function \(s(t)\). Given \(s(t)=\frac{1}{12}t^{2}-\ln(t + 1)\), using the power rule \((x^{n})^\prime=nx^{n - 1}\) and the chain rule \((\ln(u))^\prime=\frac{u^\prime}{u}\) (where \(u=t + 1\) and \(u^\prime=1\)).

$$ v(t)=s^\prime(t)=\frac{1}{12}\times2t-\frac{1}{t + 1}=\frac{t}{6}-\frac{1}{t + 1} $$

Step2: Find acceleration function

The acceleration function \(a(t)\) is the derivative of the velocity function \(v(t)\). Using the power rule and the chain rule again.

$$ a(t)=v^\prime(t)=\frac{1}{6}+\frac{1}{(t + 1)^{2}} $$

Answer:

Velocity function: \(v(t)=\frac{t}{6}-\frac{1}{t + 1}\), Acceleration function: \(a(t)=\frac{1}{6}+\frac{1}{(t + 1)^{2}}\)