4.2 Kinematics of motion in a straight line

Learning objectives

4.2.1Complex loci• understand the concepts of distance and speed as scalar quantities, and of displacement, velocity and acceleration as vector quantities Restricted to motion in one dimension only. The term 'deceleration' may sometimes be used in the context of decreasing speed.4.2.2Complex loci• sketch and interpret displacement-time graphs and velocity-time graphs, and in particular appreciate that - the area under a velocity-time graph represents displacement, - the gradient of a displacement-time graph represents velocity, - the gradient of a velocity-time graph represents acceleration 4.2.3Complex loci• use differentiation and integration with respect to time to solve simple problems concerning displacement, velocity and acceleration Calculus required is restricted to techniques from the content for Paper 1: Pure Mathematics 1.4.2.4Straight lines• use appropriate formulae for motion with constant acceleration in a straight line. Questions may involve setting up more than one equation, using information about the motion of different particles.

Quantity	Type in 1D	Meaning
distance	scalar, $\ge0$	total path length travelled
displacement $s$	signed/vector quantity	final position minus initial position
speed	scalar, $\ge0$	rate of distance; $\|v\|$
velocity $v$	signed/vector quantity	rate of displacement
acceleration $a$	signed/vector quantity	rate of velocity

Choose one positive direction before calculating. Negative displacement/velocity/acceleration means opposite to that direction, not an invalid value.

A particle moves from $x=0$ to $x=5$ m then to $x=2$ m. Distance $=5+3=8$ m; displacement $=+2$ m. If the return takes $1$ s, its velocity then is negative while speed is positive.

$Speed decreases when velocity and acceleration have opposite signs:$ va<0. $Thus negative acceleration is deceleration only while $v>0$.$

Distance is not always $|$ displacement $|$ when direction changes. “Deceleration” means decreasing speed, not simply $a<0$ .

Graph	Gradient	Area under graph	Height/sign
displacement–time	velocity	no standard motion meaning here	position relative to origin
velocity–time	acceleration	displacement	direction of motion

A secant gradient gives average velocity/acceleration over an interval; a tangent gives instantaneous value. Horizontal $s$ – $t$ means rest; horizontal $v$ – $t$ means zero acceleration.

Signed area above minus area below the time axis gives displacement. Total distance is the sum of absolute areas, splitting at every $v=0$ direction change.

A positive/negative $s$ – $t$ slope shows positive/negative velocity. On a $v$ – $t$ graph, crossing the axis may reverse direction; increasing/decreasing height alone must be interpreted with sign.

Area under an $s$ – $t$ graph is not displacement. Negative $v$ – $t$ area subtracts from displacement but adds positively to total distance.

v=\frac{ds}{dt},\qquad a=\frac{dv}{dt}=\frac{d^2s}{dt^2}; $reversing gives$ v=\int a,dt+C_1,\qquad s=\int v,dt+C_2.

Differentiate when moving right in the chain $s\to v\to a$ ; integrate when moving left. Use stated values such as $v(t_0)$ or $s(t_0)$ to determine each constant, preserving one sign convention.

$If $a=6t-2$ and $v(0)=3$, then$ v=3t^2-2t+3. $Integrating again and using $s(0)=5$ gives$ s=t^3-t^2+3t+5.

Solve $v=0$ for rest/possible direction changes; test the sign of $v$ around each root. For total distance, evaluate displacement changes separately on intervals of constant velocity sign.

Include integration constants. Calculus is restricted to Paper 1 techniques; do not import later integration methods.

$For straight-line constant acceleration:$ v=u+at,s=ut+\tfrac12at^2,v^2=u^2+2as,s=\tfrac12(u+v)t.

Choose a positive direction, assign signed $u,v,a,s$ , list known/required quantities, select a formula with no extra unknown, solve and check time/position against the event wording.

From $u=5$ m s $^{-1}$ to $v=17$ m s $^{-1}$ in $4$ s, $a=3$ m s $^{-2}$ and $s=\tfrac12(5+17)4=44$ m.

For different particles, write a separate equation for each using the same time origin and coordinate line. At a meeting, their positions are equal—not necessarily their displacements from different starting points. Other events may share time but not velocity.

These formulae require constant acceleration. Do not mix sign conventions or set two travelled distances equal unless the geometry justifies it.

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—4.3 MomentumRead

4.2 Kinematics of motion in a straight line

Learning objectives

Separate accumulated path length from signed change

Read motion graphs by gradient, signed area and zero crossings

Move between displacement, velocity and acceleration with time calculus

Select a signed constant-acceleration equation for each event

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