M1.3 - Kinematics of a particle moving in a straight line

Learning objectives

M1.3.1Constant-acceleration motion in a straight lineMotion in a straight line with Graphical solutions may be required, including constant acceleration. displacement-time, velocity-time, speed-time and acceleration-time graphs.; Knowledge and use of formulae for constant acceleration will be required.

For motion in one straight line with constant acceleration, choose one positive direction and keep every velocity, acceleration and displacement signed. Use $u$ for velocity at the start of the interval, $v$ at the end, $a$ for the constant acceleration, $s$ for signed displacement and $t\ge0$ for elapsed time. Distance and speed are non-negative magnitudes; they are not interchangeable with $s$ and $v$ .

$v=u+at,\qquad s=ut+\frac12at^2,\qquad s=vt-\frac12at^2,\qquad v^2=u^2+2as,\qquad s=\frac12(u+v)t$

Information and target	Efficient relation
$u,a,t$ known; find $v$	$v=u+at$
$u,a,t$ known; find $s$	$s=ut+\tfrac12at^2$
$v,a,t$ known; find $s$	$s=vt-\tfrac12at^2$
time absent; connect speeds and displacement	$v^2=u^2+2as$
$u,v,t$ known	$s=\tfrac12(u+v)t$

Before substituting: (1) isolate one interval in which $a$ is constant; (2) state the positive direction; (3) list the five quantities with signs and units; (4) choose a relation containing the target and no unnecessary unknown; (5) reject roots that conflict with time, direction or the described stage. For distance travelled during the $n$ th second, calculate $s(n)-s(n-1)$ within the same constant-acceleration interval.

For free vertical motion near Earth's surface, take upward as positive so $a=-g$ . A particle projected upward at $14.7\,\mathrm{m\,s^{-1}}$ reaches instantaneous rest when $0=14.7-9.8t,$ so $t=1.5$ s. Its greatest displacement above projection is $s=14.7(1.5)-\frac12(9.8)(1.5)^2=11.025\text{ m}.$ After the top, velocity is negative while speed is positive. At a return to the projection level, $s=0$ even though the total distance travelled is twice the greatest height.

Graph	Gradient	Signed area under graph	Constant- $a$ shape
displacement-time	velocity	not used as a standard motion quantity	quadratic curve; turning point where gradient is zero
velocity-time	acceleration	displacement	straight line
speed-time	rate of change of speed, not signed acceleration across a reversal	distance travelled	non-negative; a velocity sign change appears as a V-shaped turn
acceleration-time	not normally used	change in velocity	horizontal line

On a velocity-time graph, equal positions occur when the signed areas accumulated from the same starting time are equal; catching up means equal displacements, not equal velocities. On a speed-time graph, all area contributes positively to distance. Vertical segments would represent an instantaneous change and should appear only when the model explicitly permits one.

When acceleration changes, stop at the boundary and carry the boundary velocity and position into a new interval. Reset elapsed time carefully or keep one global time variable consistently. Do not apply one SUVAT equation across a bounce, string break, parachute opening or any other change of acceleration. A negative value of $v$ gives direction; report $|v|$ when speed is requested.

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—M1.4 - Dynamics of a particle moving in a straight line or planeRead