Use \(s=ut+\frac12at^2\).

So \(s=2\times4+\frac12\times3\times4^2=8+24=32\text{ m}\).

Use \(s=vt-\frac12at^2\).

So \(s=17\times8-\frac12\times2\times8^2=136-64=72\text{ m}\).

Use \(s=ut+\frac12at^2\).

So \(40=3\times5+\frac12 a\times25\).

Hence \(40=15+12.5a\), so \(25=12.5a\).

Therefore \(a=2\text{ m s}^{-2}\).

Use \(s=vt-\frac12at^2\).

So \(28=13\times4-\frac12 a\times4^2\).

Thus \(28=52-8a\), so \(8a=24\).

Hence \(a=3\text{ m s}^{-2}\).

Use \(v^2=u^2+2as\).

So \(14^2=2^2+2a(24)\).

Hence \(196=4+48a\), so \(192=48a\).

Therefore \(a=4\text{ m s}^{-2}\).

Use \(s=ut+\frac12at^2\).

So \(45=6u+\frac12\times1.5\times6^2\).

Thus \(45=6u+27\), so \(6u=18\).

Hence \(u=3\text{ m s}^{-1}\).

Use \(s=vt-\frac12at^2\).

So \(24=4v-\frac12(-2.5)\times4^2\).

Hence \(24=4v+20\), so \(4v=4\).

Therefore \(v=1\text{ m s}^{-1}\).

Use \(v^2=u^2+2as\).

So \(5^2=2^2+2(0.75)s\).

Hence \(25=4+1.5s\), so \(21=1.5s\).

Therefore \(s=14\text{ m}\).

Use \(s=ut+\frac12at^2\).

So \(24=10t+\frac12(-2)t^2=10t-t^2\).

Hence \(t^2-10t+24=0\).

Factorising gives \((t-2)(t-8)=0\).

The positive times are 2 s and 8 s, so the first time is \(2\text{ s}\).

Use \(s=vt-\frac12at^2\).

So \(21=5t-\frac12(0.5)t^2=5t-0.25t^2\).

Multiply by 4:

\(84=20t-t^2\), so \(t^2-20t+84=0\).

Factorising gives \((t-6)(t-14)=0\).

The first positive time is \(6\text{ s}\).

Use \(s=ut+\frac12at^2\).

So \(20=3t+\frac12t^2\).

Multiply by 2:

\(40=6t+t^2\), so \(t^2+6t-40=0\).

Factorising gives \((t+10)(t-4)=0\).

The first positive time is \(4\text{ s}\).

Use \(v^2=u^2+2as\).

So \(v^2=5^2+2(-2)(6)=25-24=1\).

Thus \(v=\pm1\text{ m s}^{-1}\).

Since the object has changed direction, its final velocity must be opposite in sign to the initial velocity.

Therefore \(v=-1\text{ m s}^{-1}\).

Use \(v^2=u^2+2as\).

So \(13^2=u^2+2(1)(60)\).

Hence \(169=u^2+120\), so \(u^2=49\).

Thus \(u=\pm7\text{ m s}^{-1}\).

Since the object has not changed direction, take the positive value.

Therefore \(u=7\text{ m s}^{-1}\).

Use \(v^2=u^2+2as\).

So \(v^2=3^2+2(2)(18)=9+72=81\).

Thus \(v=\pm9\text{ m s}^{-1}\).

Here the motion starts with positive velocity and has positive acceleration, so the physically relevant answer is \(9\text{ m s}^{-1}\).

The initial velocity and the acceleration are both positive. That means the particle keeps speeding up in the same direction, so the negative root from \(v^2=81\) is not physically relevant.

Use \(v^2=u^2+2as\).

Here \(v=0\), \(u=20\), \(a=-4\).

So \(0=20^2+2(-4)s=400-8s\).

Hence \(8s=400\), so \(s=50\text{ m}\).

Use \(v^2=u^2+2as\).

So \(60^2=0+2a(400)\).

Hence \(3600=800a\), so \(a=4.5\text{ m s}^{-2}\).

Use \(v^2=u^2+2as\).

So \(80^2=0+2(4)s\).

Hence \(6400=8s\), so \(s=800\text{ m}\).

Use \(v^2=u^2+2as\).

So \(0=20^2+2a(40)=400+80a\).

Hence \(80a=-400\), so \(a=-5\text{ m s}^{-2}\).

The deceleration is therefore \(5\text{ m s}^{-2}\).

Use \(s=\frac{u+v}{2}t\).

So \(240=\frac{30+v}{2}\times 12\).

Hence \(240=6(30+v)\).

So \(40=30+v\), giving \(v=10\text{ m s}^{-1}\).

The solution uses constant-acceleration equations, so we assume the car moves in a straight line with uniform deceleration over the 12 s, and we ignore effects such as changing slope or varying resistance.