M2.1 - Kinematics of a particle moving in a straight line or plane

Learning objectives

M2.1.1Motion in a vertical plane with constant acceleration, eMotion in a vertical plane with constant acceleration, e.g. under gravity.M2.1.2Simple cases of motion of a projectileSimple cases of motion of a projectile.M2.1.3Velocity and acceleration whenVelocity and acceleration when the The setting up and solution of equations of the form displacement is a function of time. dx dv = f(t) or = g(t) will be consistent with the level of dt dt calculus in P1, P2 P3 and P4.M2.1.4Differentiating and integrating vectorsDifferentiate and integrate vector-valued position functions with respect to time to obtain velocity and acceleration.

Motion in a vertical plane is two-dimensional, but a constant acceleration lets each perpendicular component be modelled with the same signed SUVAT relationships. Choose fixed unit vectors $\mathbf i$ horizontally and $\mathbf j$ vertically, then keep one common time $t$ for both components.

$\mathbf v=\mathbf u+\mathbf a t,\qquad \mathbf r=\mathbf r_0+\mathbf u t+\frac12\mathbf a t^2$

For free motion under gravity with upward $\mathbf j$ , acceleration is constant and vertical: $\mathbf a=-g\mathbf j$ . Hence horizontal velocity remains constant, while vertical velocity changes by $-gt$ . Gravity changes neither the horizontal component directly nor the chosen coordinate axes.

Component	Velocity	Displacement from the initial point
horizontal	$v_x=u_x+a_xt$	$x=u_xt+\tfrac12a_xt^2$
vertical	$v_y=u_y+a_yt$	$y=u_yt+\tfrac12a_yt^2$
free motion under gravity	$v_x=u_x$ , $v_y=u_y-gt$	$x=u_xt$ , $y=u_yt-\tfrac12gt^2$

A particle starts at the origin with velocity $(6\mathbf i+8\mathbf j)\,\mathrm{m\,s^{-1}}$ and acceleration $-9.8\mathbf j\,\mathrm{m\,s^{-2}}$ . After $1$ s, $\mathbf r=6\mathbf i+(8-4.9)\mathbf j=6\mathbf i+3.1\mathbf j\text{ m},$ and $\mathbf v=6\mathbf i-1.8\mathbf j\,\mathrm{m\,s^{-1}}$ . The negative vertical velocity means downward motion; the particle may still be above its starting level.

Do not apply one scalar SUVAT equation to a vector without resolving it, use different times for the two components, or take $g$ as positive regardless of the chosen upward axis. This card establishes the constant-acceleration framework; launch, flight and trajectory decisions are developed in the next objective.

A projectile is modelled as a particle that, after projection, moves freely under gravity. With air resistance neglected, its horizontal velocity is constant and its vertical acceleration is $-g$ when upward is positive. The two component motions are independent but describe the same particle at the same time.

$u_x=u\cos\alpha,\qquad u_y=u\sin\alpha$

$x=(u\cos\alpha)t,\qquad y=(u\sin\alpha)t-\frac12gt^2,\qquad \mathbf v=(u\cos\alpha)\mathbf i+(u\sin\alpha-gt)\mathbf j$

Use the component whose displacement condition is known to find the physically relevant $t\ge0$ , then substitute that same time into the other component. At greatest height $v_y=0$ , not the whole velocity. At impact, use $|\mathbf v|=\sqrt{v_x^2+v_y^2}$ and obtain its direction from the signed components.

$y=x\tan\alpha-\frac{gx^2}{2u^2\cos^2\alpha}=x\tan\alpha-\frac{gx^2}{2u^2}(1+\tan^2\alpha)$

A particle is projected at $20\,\mathrm{m\,s^{-1}}$ at $30^\circ$ above horizontal ground and lands at its launch level. From $0=10t-4.9t^2$ , the non-zero flight time is $t=2.04$ s. Its horizontal range is $(20\cos30^\circ)(2.04)=35.3$ m, and its greatest height is found from $0^2=10^2-2(9.8)h$ , giving $h=5.10$ m.

A negative root or the root $t=0$ may describe the wrong event; retain only times consistent with the stated flight. Do not assume equal launch and landing heights unless given, set the entire velocity to zero at the top, or use the trajectory equation before defining the origin and angle.

When signed displacement $x$ is a differentiable function of time, its first derivative is velocity and its second derivative is acceleration. Differentiation reveals instantaneous motion; integration reconstructs velocity or displacement only after the relevant initial condition fixes the constant.

$v=\frac{dx}{dt},\qquad a=\frac{dv}{dt}=\frac{d^2x}{dt^2}$

$v=\int a\,dt+C_1,\qquad x=\int v\,dt+C_2$

Mathematical result	Motion meaning
$v(t)=0$	instantaneous rest; test the sign on either side to see whether direction changes
$v(t)>0$ or $v(t)<0$	motion in the positive or negative chosen direction
$a(t)$ has a sign	velocity is increasing or decreasing, not necessarily speed
$\|x(t_2)-x(t_1)\|$	distance only if no reversal occurs inside the interval
several direction intervals	total distance is the sum of absolute displacement changes

For $t\ge0$ , let $x=t^3-6t^2+9t$ metres. Then $v=3(t-1)(t-3),\qquad a=6t-12.$ The particle is instantaneously at rest at $t=1$ and $t=3$ , and the velocity changes sign at both times. Since $x(0)=0$ , $x(1)=4$ , $x(3)=0$ and $x(4)=4$ , the distance travelled from $t=0$ to $t=4$ is $4+4+4=12$ m, not $|x(4)-x(0)|=4$ m.

Do not divide $x(t)$ by $t$ to obtain instantaneous velocity, omit constants after integration, or treat signed displacement as total distance across a reversal. This objective is one-dimensional; vector position and componentwise calculus follow next.

A position vector $\mathbf r(t)=x(t)\mathbf i+y(t)\mathbf j$ records a particle's coordinates relative to a fixed origin. Differentiate or integrate each component with respect to the same time variable; the constant of integration is itself a vector fixed by a position or velocity condition.

$\mathbf v=\frac{d\mathbf r}{dt}=\dot x\mathbf i+\dot y\mathbf j,\qquad \mathbf a=\frac{d\mathbf v}{dt}=\frac{d^2\mathbf r}{dt^2}$

$\mathbf v=\int\mathbf a\,dt+\mathbf C,\qquad \mathbf r=\int\mathbf v\,dt+\mathbf D$

Motion statement	Component condition
instantaneous rest	every component of $\mathbf v$ is zero at the same time
moving parallel to $\mathbf i$	the $\mathbf j$ -component of velocity is zero
moving parallel to $a\mathbf i+b\mathbf j$	velocity components are in the ratio $a:b$ , with direction checked
speed	$\|\mathbf v\|=\sqrt{v_x^2+v_y^2}$
distance from the origin	$\|\mathbf r\|=\sqrt{x^2+y^2}$

Suppose $\mathbf v=(3t^2-3)\mathbf i+2t\mathbf j\quad\text{and}\quad\mathbf r(0)=2\mathbf i-\mathbf j.$ Integrating and applying the initial position gives $\mathbf r=(t^3-3t+2)\mathbf i+(t^2-1)\mathbf j.$ Also $\mathbf a=6t\mathbf i+2\mathbf j$ . At $t=1$ , $\mathbf v=2\mathbf j\,\mathrm{m\,s^{-1}}$ , so the motion is parallel to $\mathbf j$ and the speed is $2\,\mathrm{m\,s^{-1}}$ .

A zero component does not mean the entire vector is zero, and equal component ratios may describe the opposite direction if their common multiplier is negative. Do not replace vector integration constants with one scalar constant or confuse $|\mathbf r|$ with distance travelled along a curved path.

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M2.1 - Kinematics of a particle moving in a straight line or plane

Learning objectives

Model constant-acceleration motion in a vertical plane

Resolve a projectile into one shared flight

Read one-dimensional motion from displacement functions

Differentiate and integrate motion vectors componentwise

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