Answer: speed \(=\frac3{16}\); the acceleration is always positive.

Answer: speed \(=\frac3{16}\); the acceleration is always positive.

Differentiate \(s=(1+3t)^{-1/2}\):

\(\frac{ds}{dt}=-\frac32(1+3t)^{-3/2}.\)

When \(t=1\),

Speed is the magnitude of velocity, so the speed is \(\frac3{16}\).

Differentiate again:

For \(t\ge0\), \(1+3t\gt 0\), so \((1+3t)^{-5/2}\gt 0\). Hence the acceleration is positive and can never be zero.