Motion in a Straight Line

$u$ = velocity at $t=0$, $v$ = velocity at time $t$. Hold only while $a$ is truly constant over the interval — re-derive (or split into constant-$a$ pieces) otherwise.

Displacement: $\Delta x = x_2 - x_1$, signed, net change only. Distance (path length): always positive, only ever adds up.

1. No reversal: distance = |displacement|.
2. Velocity reverses: distance = sum of |Δx| each leg; displacement = net Δx. Distance ≥ |displacement|, always.

Average speed = path length / $\Delta t$ (not displacement / $\Delta t$). Always $\text{avg speed} \ge |\bar v|$.

1. Round trip (ends where it started): displacement = 0, so $\bar v = 0$ — even though average speed is clearly nonzero.

Slope of the tangent to the $x\text{-}t$ graph. Positive while $x$ increases, negative while decreasing, zero at a momentary flat (e.g. top of a vertical throw).

1. Instantaneous speed = $|v|$ — exactly, unlike the averaged versions, since over a vanishing interval the path can’t double back on itself.

SI unit: $\text{m/s}^2$. Geometrically, slope of the $v\text{-}t$ graph.

1. Same sign as $v$ (both + or both −): speed is increasing, regardless of whether $a$ itself reads positive or negative.
2. Opposite sign to $v$: speed is decreasing — this is true deceleration.
3. Never judge from the sign of $a$ alone — always compare it against the sign of $v$.

Apply the same sign convention consistently to both $v_A$ and $v_B$.

1. Same direction (chasing): relative speed is reduced — A appears slower to B than its ground speed.
2. Opposite directions (closing in): relative speed adds — A appears to rush past much faster than its ground speed.

• Slope of $x\text{-}t$ graph = $v$ (instantaneous velocity).
• Slope of $v\text{-}t$ graph = $a$ (acceleration).
• Area under $v\text{-}t$ graph, between two instants = displacement $s$ over that interval.