A particle moves along x-axis with position $x = 2t^2 - 3t + 1$ (in metres, $t$ in seconds). The average speed of the particle in the time interval $t = 0$ to $t = 3$ s is:

Uniform acceleration (SUVAT): $v = u + at$, $s = ut + \frac{1}{2}at^2$, $v^2 = u^2 + 2as$, $s = \frac{u+v}{2}t$. Distance in $n$th second: $s_n = u + \frac{a}{2}(2n-1)$. Non-uniform: use calculus: $v = dx/dt$, $a = dv/dt = v\,dv/dx$.

Displacement-time: slope = velocity. Straight line = uniform velocity. Curve = non-uniform velocity. Velocity-time: slope = acceleration. Area under curve = displacement. Trapezoid under v-t graph = displacement for uniform acceleration.

Horizontal: $x = v_0\cos\theta \cdot t$ (uniform velocity). Vertical: $y = v_0\sin\theta \cdot t - \frac{1}{2}gt^2$ (uniform acceleration $-g$). Time of flight: $T = 2v_0\sin\theta/g$. Range: $R = v_0^2\sin 2\theta/g$ (max at $\theta = 45^{\circ}$). Max height: $H = v_0^2\sin^2\theta/2g$.

Relative velocity: $\vec{v}_{AB} = \vec{v}_A - \vec{v}_B$. If $v_A$ and $v_B$ in same direction: $v_{rel} = v_A - v_B$. Opposite: $v_{rel} = v_A + v_B$. Applications: river crossing (to go straight perpendicular: aim upstream at angle $\sin\theta = v_{river}/v_{boat}$). Rain problem: use relative velocity to find angle to hold umbrella.

On a position-time graph the slope gives velocity, and a curved line therefore means changing velocity. On a velocity-time graph the slope gives acceleration and the area under the curve gives displacement, with area below the axis counting as negative. These two facts answer most graph questions without a single formula. A useful consistency check: wherever the position-time graph is steepest, the velocity-time graph must be at its highest point.

The velocity of A with respect to B is the vector difference $\vec{v}_{AB} = \vec{v}_A - \vec{v}_B$, which means subtracting components, not magnitudes. The standard exam scenario is rain falling vertically at $v_r$ while a person walks horizontally at $v_m$: the rain appears to come at an angle $\tan\theta = v_m/v_r$ from the vertical, which is why an umbrella must be tilted forward. For a river crossing, the shortest path and the shortest time require different headings, and a question will always specify which one it wants.

Choose one direction as positive and keep it for the whole problem. Taking upward as positive makes $a = -g$ throughout, including on the way down, and the equations then handle the rise and fall in a single calculation without splitting the motion into two parts. Two symmetry results follow immediately and are worth using directly: the time to rise equals the time to fall for the same height, and the speed at any height on the way down equals the speed at that height on the way up. A body thrown up with speed u returns to the launch point with speed u, in time $2u/g$.