Constant acceleration (suvat) equationsAQA A-Level Maths: Revision notes

The suvat equations describe motion in a straight line with constant acceleration. The five variables are:

• $s$ displacement, $u$ initial velocity, $v$ final velocity, $a$ acceleration, $t$ time. In any problem you will know three of these and need a fourth. The fifth is not needed, so choose the equation that does not contain it. Choose a positive direction and give every vector ($s$, $u$, $v$, $a$) the correct sign. If the acceleration changes, apply the equations separately to each stage.

$v=u+at\qquad s=\tfrac12(u+v)t\qquad s=ut+\tfrac12at^2$ $v^2=u^2+2as\qquad s=vt-\tfrac12at^2$ Each one leaves out one variable:

• $v=u+at$ has no $s$
• $s=\frac12(u+v)t$ has no $a$
• $s=ut+\frac12at^2$ has no $v$
• $v^2=u^2+2as$ has no $t$
• $s=vt-\frac12at^2$ has no $u$. Learn all five; they are derived below.

Start from the definition of acceleration as the rate of change of velocity: $a=\frac{v-u}{t}$, so $v=u+at.$ For constant acceleration the average velocity is $\frac12(u+v)$, so the displacement is $s=\frac12(u+v)t.$ Substitute $v=u+at$: $s=\frac12(2u+at)t=ut+\frac12at^2.$ To remove $t$, use $t=\frac{v-u}{a}$ in $s=\frac12(u+v)t$: $s=\frac{v^2-u^2}{2a}$, so $v^2=u^2+2as.$ These can also be derived from a velocity-time graph: the area of the trapezium with parallel sides $u$ and $v$ and width $t$ is $\frac12(u+v)t$, and the gradient is $a$.

Example: a car's speed rises from $8$ m s$^{-1}$ to $20$ m s$^{-1}$ in $6$ s.

• Acceleration: $a=\frac{v-u}{t}=\frac{20-8}{6}=2$ m s$^{-2}$.
• Distance: $s=\frac12(8+20)(6)=84$ m.
• Speed after $50$ m: $v^2=8^2+2(2)(50)=264$, so $v=16.2$ m s$^{-1}$. Example with deceleration: a train at $30$ m s$^{-1}$ stops in $450$ m. Using $0=30^2+2a(450)$ gives $a=-1$ m s$^{-2}$. The negative sign shows that it is slowing down. Quadratic equations, such as $s=ut+\frac12at^2$ solved for $t$, can give two values: reject any that do not fit the situation.

When two objects move, write a displacement for each using the same time $t$. They meet or one overtakes the other when their displacements from the same point are equal. Example: car $A$ has $u=10$, $a=1.5$; car $B$ starts from rest with $a=2.5$, both from $P$. $s_A=10t+0.75t^2$ and $s_B=1.25t^2$. Equating, $10t=0.5t^2$, so $t=20$ s (not $t=0$), and the distance from $P$ is $1.25\times400=500$ m. When the two velocities are equal ($10+1.5t=2.5t$, $t=10$ s) the cars are $175-125=50$ m apart.