Answer: \(v=\dfrac{(2t-3)^3}{6}+\dfrac32\), \(t=\dfrac{3-\sqrt[3]{9}}{2}\), displacement \(=\dfrac{293}{48}\), and approximate change in \(v\) is \(0.08\).

(a)(i) Since acceleration is the derivative of velocity,

\(\dfrac{\mathrm dv}{\mathrm dt}=(2t-3)^2.\)

Integrate:

\(v=\int(2t-3)^2\,\mathrm dt.\)

Let \(u=2t-3\). Then \(\mathrm du=2\,\mathrm dt\), so

\(v=\dfrac{(2t-3)^3}{6}+c.\)

Use the condition \(v=6\) when \(t=3\):

\(6=\dfrac{(2(3)-3)^3}{6}+c=\dfrac{27}{6}+c.\)

Therefore

\(c=\dfrac32.\)

Hence

\(v=\dfrac{(2t-3)^3}{6}+\dfrac32.\)

(a)(ii) The particle is at rest when \(v=0\):

\(\dfrac{(2t-3)^3}{6}+\dfrac32=0.\)

Multiplying by \(6\),

\((2t-3)^3+9=0.\)

Thus

\((2t-3)^3=-9,\)

so

\(2t-3=-\sqrt[3]{9}.\)

Therefore

\(t=\dfrac{3-\sqrt[3]{9}}2.\)

(a)(iii) Displacement is found by integrating velocity:

\(s=\int\left(\dfrac{(2t-3)^3}{6}+\dfrac32\right)\,\mathrm dt.\)

This gives

\(s=\dfrac{(2t-3)^4}{48}+\dfrac32t+d.\)

Use the condition \(s=4\) when \(t=\dfrac52\):

\(4=\dfrac{(5-3)^4}{48}+\dfrac32\cdot\dfrac52+d.\)

So

\(4=\dfrac{16}{48}+\dfrac{15}{4}+d=\dfrac13+\dfrac{15}{4}+d.\)

Therefore

\(d=-\dfrac1{12}.\)

Thus

\(s=\dfrac{(2t-3)^4}{48}+\dfrac32t-\dfrac1{12}.\)

When \(t=3\),

\(s=\dfrac{3^4}{48}+\dfrac92-\dfrac1{12}.\)

Using denominator \(48\),

\(s=\dfrac{81}{48}+\dfrac{216}{48}-\dfrac4{48}=\dfrac{293}{48}.\)

(b) For a small increase \(\Delta t\),

\(\Delta v\approx\dfrac{\mathrm dv}{\mathrm dt}\Delta t.\)

But \(\dfrac{\mathrm dv}{\mathrm dt}=a=(2t-3)^2\). At \(t=\dfrac52\),

\(a=(5-3)^2=4.\)

With \(\Delta t=0.02\),

\(\Delta v\approx4(0.02)=0.08.\)