Constant acceleration formulaeEdexcel International A Level Maths: Revision notes

Model the object as a particle moving in a straight line with constant (uniform) acceleration. Five quantities describe the motion:

• $s$: displacement from the starting point (m)
• $u$: initial velocity (m s⁻¹)
• $v$: final velocity (m s⁻¹)
• $a$: acceleration (m s⁻²)
• $t$: time (s) All of $s$, $u$, $v$, $a$ are vectors along the line, so choose a positive direction and give each quantity a sign. A deceleration is a negative $a$ in the positive direction of motion. The formulae only apply while $a$ is constant, so a journey with changing acceleration must be split into stages.

These must be known: $v=u+at\qquad s=ut+\tfrac12at^2\qquad s=vt-\tfrac12at^2$ $v^2=u^2+2as\qquad s=\tfrac12(u+v)t$ The first comes from the definition of acceleration, $a=\frac{v-u}{t}$. The last uses the average velocity $\frac12(u+v)$, which is valid because the acceleration is constant. Each formula leaves out exactly one of the five variables:

• no $s$: $v=u+at$
• no $v$: $s=ut+\frac12at^2$
• no $u$: $s=vt-\frac12at^2$
• no $t$: $v^2=u^2+2as$
• no $a$: $s=\frac12(u+v)t$

Write down what you are given, what you want, and the variable you are not involved with. Pick the formula that leaves out that variable. Example: a cyclist passes a point at $4$ m s⁻¹ and accelerates at $0.5$ m s⁻² for $12$ s. Find the distance. Known: $u=4$, $a=0.5$, $t=12$. Wanted: $s$. Not involved: $v$. So $s=ut+\frac12at^2=48+36=84$ m. A quadratic in $t$ can arise from $s=ut+\frac12at^2$ when $t$ is the unknown. Solve it and reject any negative or physically impossible root. Always give final answers to 2 or 3 significant figures unless an exact value is asked for.

An object moving freely under gravity has constant acceleration $g=9.8$ m s⁻² downwards (Edexcel IAL uses $g=9.8$). Take up as positive and use $a=-9.8$.

• At the highest point the velocity is $0$.
• Time up equals time down to the same level, and the speed is the same at the same height.
• Landing below the starting point gives a negative displacement $s$. Example: a ball thrown up at $14$ m s⁻¹ from $2$ m above ground. Rise: $0=14^2-2(9.8)h$ gives $h=10$ m, so the greatest height above ground is $12$ m. Landing speed: $v^2=14^2+2(9.8)(2)=235.2$ gives $15.3$ m s⁻¹. The model assumes no air resistance and that the object is a particle.

If the acceleration changes, split the motion into stages. The final velocity of one stage is the initial velocity of the next. Example: a train accelerates at $0.5$ m s⁻² for $40$ s from rest to $20$ m s⁻¹, travels at $20$ m s⁻¹ for $150$ s (distance $3000$ m) then decelerates at $0.8$ m s⁻². From $0=20^2-2(0.8)s$ the braking distance is $250$ m. For two moving particles, write each displacement in terms of the same time $t$ (measured from the same instant). "Meets" or "overtakes" means equal displacements from a common point. The gap between them is greatest or least when their speeds are equal. If a particle stops accelerating at some time, use different expressions before and after that time.