Why Doubling Your Speed Makes Braking Distance Four Times Longer

When acceleration stays constant, five quantities describe the motion: **initial velocity $u$, final velocity $v$, acceleration $a$, time $t$ and displacement $s$.** Three equations link them, and knowing any three quantities gives the other two.

This part covers deriving the equations, free fall and stopping distance, relative velocity, and overtaking and meeting problems.

**For constant acceleration, the slope and area of a straight v-t graph — or integrating $a=\frac{dv}{dt}$ and $v=\frac{dx}{dt}$ — give $v=u+at$, $s=ut+\frac{1}{2}a{t}^2$ and $v^2=u^2+2as$.

Graphical method.** The v-t graph is a straight line from $u$ to $v$ over time $t$.

- Slope $a=\frac{v-u}{t}$, so $v=u+at$
- Area = rectangle $ut$ + triangle $\frac{1}{2}(at)t$, so $s=ut+\frac{1}{2}a{t}^2$
- Area = trapezium $\frac{1}{2}(u+v)t$ with $t=\frac{v-u}{a}$, so $s=\frac{v^2-u^2}{2a}$

Calculus method.

$${\int }_u^{v}dv={\int }_0^{t}adt\Rightarrow v=u+at{\int }_0^{s}dx={\int }_0^t(u+at)dt\Rightarrow s=ut+\frac{1}{2}a{t}^2$$

$$a=v\frac{dv}{dx}\Rightarrow {\int }_u^{v}vdv={\int }_0^{s}adx\Rightarrow v^2=u^2+2as$$

Worked example. A scooter moving at $10$ m/s accelerates at $2$ m/s${}^2$ for $5$ s.

$$v=10+2(5)=20\text{ m/s}s=10(5)+\frac{1}{2}(2)(25)=75\text{ m}$$

Check: $v^2=100+2(2)(75)=400$, so $v=20$ m/s.

An everyday example. A scooter pulling away from a traffic signal follows these equations while its acceleration stays steady.

The substance. These equations fail when acceleration changes — then you must integrate directly.

**Free fall is uniformly accelerated motion with $a=g$ downward, and a braking vehicle is uniformly retarded motion, so both use the kinematic equations with a consistent sign convention.

Take upward as positive** and $g=10$ m/s${}^2$.

Worked example 1 — thrown up. A ball is thrown up at $20$ m/s.

$$\text{time to top: }0=20-10t\Rightarrow t=2\text{ s}\text{height: }h=\frac{u^2}{2g}=\frac{400}{20}=20\text{ m}$$

Worked example 2 — dropped. A stone falls from rest through $45$ m.

$$45=\frac{1}{2}(10)t^2\Rightarrow t=3\text{ s}v=10\times 3=30\text{ m/s}$$

Worked example 3 — stopping distance. A car at $72$ km/h $=20$ m/s brakes at $5$ m/s${}^2$. Suppose the driver's reaction time is $0.5$ s.

$$\text{reaction distance}=20\times 0.5=10\text{ m}\text{braking distance}=\frac{u^2}{2a}=\frac{400}{10}=40\text{ m}$$

Total stopping distance $=50$ m. At $40$ m/s, the braking distance becomes $\frac{1600}{10}=160$ m.

An everyday example. Low speed limits near schools make sense because braking distance grows with the square of speed.

The substance. **At the highest point, velocity is zero but acceleration is still $g$** downward.

**The velocity of A relative to B is $v_{AB}=v_A-v_B$, which is how fast and in which direction A appears to move to an observer riding with B.**

Similarly, relative position is $x_{AB}=x_A-x_B$, and relative acceleration is $a_{AB}=a_A-a_B$.

Worked example 1 — same direction. Train A runs east at $72$ km/h $=20$ m/s and train B east at $54$ km/h $=15$ m/s.

$$v_{AB}=20-15=5\text{ m/s east}{v}_{BA}=-5\text{ m/s}$$

Worked example 2 — opposite directions. If B runs west, $v_B=-15$ m/s:

$$v_{AB}=20-(-15)=35\text{ m/s east}$$

An everyday example. Sitting in a train at a station, a train on the next track moving slowly the same way seems to crawl, while one coming the other way flashes past.

The substance. **$v_{AB}=-v_{BA}$** — the same speed, seen in the opposite direction.

Sit in the frame of one body, so it is at rest; the other body then moves with the relative velocity and relative acceleration, and only the relative gap has to be covered.

Worked example 1 — overtaking. A $200$ m train at $20$ m/s overtakes a $100$ m train at $15$ m/s on a parallel track. To pass completely, the relative distance is $200+100=300$ m.

$$t=\frac{300}{20-15}=60\text{ s}\text{ground distance by faster train}=20\times 60=1200\text{ m}$$

Worked example 2 — meeting. Two cyclists $1$ km apart ride towards each other at $15$ m/s and $10$ m/s.

$$t=\frac{1000}{15+10}=40\text{ s}\text{meeting point}=15\times 40=600\text{ m from the first}$$

Worked example 3 — with acceleration. A car passes a parked jeep at a steady $20$ m/s; at that instant the jeep starts from rest at $2$ m/s${}^2$. In the car's frame, the jeep starts at $-20$ m/s with acceleration $2$ m/s${}^2$:

$$0=-20t+\frac{1}{2}(2)t^2\Rightarrow t=20\text{ s}$$

The jeep catches up after $20$ s and $400$ m, moving at $40$ m/s.

An everyday example. Overtaking a truck on a highway takes long when your speeds are close, because only the small relative speed eats up the gap.

The substance. If both bodies have the same acceleration, their relative motion is uniform, which is why two objects dropped together stay the same distance apart.