A lion is watching a zebra from $35 \mathrm{m}$ behind it. Both are stationary. The lion then starts chasing by accelerating at a constant rate of $3 \mathrm{m}{ \mathrm{s}}^{-2}$ for $5 \mathrm{s}.$ Once at top speed the lion maintains that speed for $3 \mathrm{s}$ before decelerating at $0.5 \mathrm{m}{ \mathrm{s}}^{-2}.$ The zebra starts moving $1 \mathrm{s}$ after the lion started, accelerating at a constant rate of $2 \mathrm{m}{ \mathrm{s}}^{-2}$ for $7 \mathrm{s}$ before maintaining a constant speed.

Show that the gap between them at time $t \mathrm{s},$ for $t>8,$ after the start of the lion's motion is given by $\frac{1}{4}{t}^2-5t+\frac{51}{2}$ and, hence, determine when the lion catches the zebra, or when the lion gets closest and how close it gets.

A car is behind a tractor on a single-lane straight road with one lane in each direction. Both are moving at $15 \mathrm{m}{ \mathrm{s}}^{-1}.$ The speed limit is $25 \mathrm{m}{ \mathrm{s}}^{-1},$ so the car wants to overtake. The safe distance between the car and the tractor is $20 \mathrm{m}.$

To overtake, the car goes onto the other side of the road and accelerates at a constant rate of $2 \mathrm{m}{ \mathrm{s}}^{-2}$ until reaching the speed limit, when it continues at constant speed. Show that the distance the car is ahead of the tractor at time $t s$ after it starts to accelerate is given by $t^2-20$ for $0<t⩽5,$ and deduce that the car is not a safe distance ahead of the tractor before reaching the speed limit.

A car is behind a tractor on a single-lane straight road with one lane in each direction. Both are moving at $15 \mathrm{m}{ \mathrm{s}}^{-1}.$ The speed limit is $25 \mathrm{m}{ \mathrm{s}}^{-1},$ so the car wants to overtake. The safe distance between the car and the tractor is $20 \mathrm{m}.$

The car pulls in ahead of the tractor once it is a safe distance ahead. Find the total time taken from the start of the overtaking manoeuvre until the car has safely overtaken the tractor.

A car is behind a tractor on a single-lane straight road with one lane in each direction. Both are moving at $15 \mathrm{m}{ \mathrm{s}}^{-1}.$ The speed limit is $25 \mathrm{m}{ \mathrm{s}}^{-1},$ so the car wants to overtake. The safe distance between the car and the tractor is $20 \mathrm{m}.$

To overtake safely on the single-lane road, when the car returns to the correct side of the road in front of the tractor there must be a gap between the car and oncoming traffic of at least $20 \mathrm{m}.$ Assuming a car travelling in the opposite direction is moving at the speed limit, find the minimum distance it must be from the initial position of the overtaking car at the point at which it starts to overtake.

The sketch shows a velocity-time graph for a skier going down a slope. Given that the skier covers $80 \mathrm{m}$ during the first stage of acceleration, find the total distance covered.

The diagram shows the velocity-time graph for a particle $P$ which travels on a straight line $AB,$ where $v \mathrm{m}{ \mathrm{s}}^{-1}$ is the velocity of $P$ at time $t s.$ The graph consists of five straight line segments. The particle starts from rest when $t=0$ at a point $X$ on the line between $A$ and $B$ and moves towards $A.$ The particle comes to rest at $A$ when $t=2.5.$

Given that the distance $XA$ is $4 \mathrm{m},$ find the greatest speed reached by $P$ during this stage of the motion.

The diagram shows the velocity-time graph for a particle $P$ which travels on a straight line $AB,$ where $v \mathrm{m}{ \mathrm{s}}^{-1}$ is the velocity of $P$ at time $t s.$ The graph consists of five straight line segments. The particle starts from rest when $t=0$ at a point $X$ on the line between $A$ and $B$ and moves towards $A.$ The particle comes to rest at $A$ when $t=2.5.$

In the second stage, $P$ starts from rest at $A$ when $t=2.5$ and moves towards $B.$ The distance $AB$ is $48 \mathrm{m}.$ The particle takes $12 \mathrm{s}$ to travel from $A$ to $B$ and comes to rest at $B.$ For the first $2 \mathrm{s}$ of this stage $P$ accelerates at $3 \mathrm{m}{ \mathrm{s}}^{-2},$ reaching a velocity of $V \mathrm{m}{ \mathrm{s}}^{-1}.$ Find

the value of $V,$