For constant acceleration $a$ in a straight line:
| Equation | Form |
|---|---|
| $v = u + at$ | Velocity-time |
| $s = ut + \frac{1}{2}at^2$ | Displacement-time |
| $v^2 = u^2 + 2as$ | Velocity-displacement |
| $s = \frac{u+v}{2}\cdot t$ | Average velocity |
Derivation from $a = dv/dt$:
$v = \int a\,dt = at + C$. At $t=0$: $v = u$, so $C = u$. Thus $v = u + at$. \checkmark
$s = \int v\,dt = \int(u+at)\,dt = ut + \frac{1}{2}at^2 + C_2$. At $t=0$: $s=0$ (taking origin at initial position), so $C_2=0$. \checkmark
Choosing the right equation: Use the equation that contains the three known quantities and the one unknown.
Example 1: A ball is thrown upward at 20 m/s. Taking $a = -9.8$ m/s$^2$ (upward positive).
Time to reach maximum height: \$0 = 20 - 9.8t \Rightarrow t = 20/9.8 \approx 2.04$ s.
Maximum height: $v^2 = u^2 + 2as \Rightarrow 0 = 400 - 19.6s \Rightarrow s \approx 20.4$ m.
Apply suvat equations independently in each component:
$$\mathbf{v} = \mathbf{u} + \mathbf{a}t$$
$$\mathbf{r} = \mathbf{r}_0 + \mathbf{u}t + \tfrac{1}{2}\mathbf{a}t^2$$
In components:
$$x = x_0 + u_x t + \tfrac{1}{2}a_x t^2, \qquad y = y_0 + u_y t + \tfrac{1}{2}a_y t^2$$
Example 2: A projectile is launched from the origin at $30$ m/s at $45^\circ$ above horizontal ($g = 9.8$ m/s$^2$).
$u_x = 30\cos45^\circ = 15\sqrt{2}$ m/s, $u_y = 30\sin45^\circ = 15\sqrt{2}$ m/s, $a_x = 0$, $a_y = -9.8$ m/s$^2$.
$$x = 15\sqrt{2}\,t, \qquad y = 15\sqrt{2}\,t - 4.9t^2$$
Time of flight (when $y = 0$, $t > 0$): $t(15\sqrt{2} - 4.9t) = 0 \Rightarrow t = \dfrac{15\sqrt{2}}{4.9} \approx 4.33$ s.
Range: $x = 15\sqrt{2} \times 4.33 \approx 91.8$ m.
The suvat equations are special cases of the general integrations:
$$v(t) = u + \int_0^t a\,d\tau = u + at \quad (a \text{ constant})$$
$$s(t) = \int_0^t v\,d\tau = ut + \frac{a t^2}{2}$$
When acceleration is not constant, these integrals must be evaluated for the actual $a(t)$.
KEY TAKEAWAY: The four suvat equations apply only when acceleration is constant. In 2D, decompose into $x$ and $y$ components and apply them independently in each direction.
EXAM TIP: Always define a positive direction at the start of a kinematics question. Upward is conventionally positive for vertical motion; rightward for horizontal.
COMMON MISTAKE: Using $g = 9.8$ m/s$^2$ as positive in $v = u + at$ when upward has been taken as positive. If upward is positive, the acceleration due to gravity is $-9.8$ m/s$^2$.
