Answer: (a) \(v=\frac{45}{t+1}+5\); (b) \(122\text{ m}\) approximately.

Answer: (a) \(v=\frac{45}{t+1}+5\); (b) \(122\text{ m}\) approximately.

(a) Since acceleration is the derivative of velocity,

\(\frac{dv}{dt}=-\frac{45}{(t+1)^2}.\)

Integrate with respect to \(t\):

\(v=\int -45(t+1)^{-2}\,dt.\)

This gives

\(v=45(t+1)^{-1}+C.\)

So

\(v=\frac{45}{t+1}+C.\)

When \(t=0\), \(v=50\). Therefore

\(50=45+C,\)

so \(C=5\). Hence

\(v=\frac{45}{t+1}+5.\)

(b) Distance travelled is the integral of velocity:

An antiderivative is

\(45\ln(t+1)+5t.\)

Therefore the distance is

So

\(=45\ln11+50-45\ln2-5.\)

Hence

\(=45\ln\frac{11}{2}+45.\)

This is

\(121.7\ldots\)

so the distance travelled is

\(122\text{ m}\)

to 3 significant figures.