A-Level Maths OCR A H240

3.02: Kinematics

#3.02a

Understand and be able to use the language of kinematics: position, displacement, distance, distance travelled, velocity, speed, acceleration, equation of motion.

Learners should understand the vector nature of displacement, velocity and acceleration and the scalar nature of distance travelled and speed.

#3.02b

Understand, use and interpret graphs in kinematics for motion in a straight line.

#3.02c

Be able to interpret displacement-time and velocity-time graphs, and in particular understand and be able to use the facts that the gradient of a displacement-time graph represents the velocity, the gradient of a velocity-time graph represents the acceleration, and the area between the graph and the time axis for a velocity-time graph represents the displacement.

#3.02d

Understand, use and derive the formulae for constant acceleration for motion in a straight line: v=u+atv = u + ats=ut+12at2s = ut + \frac{1}{2}at^2s=12(u+v)ts = \frac{1}{2}(u+v)tv2=u2+2asv^2 = u^2 + 2ass=vt12at2s = vt - \frac{1}{2}at^2

*Learners may be required to derive the constant acceleration formulae using a variety of techniques:

  1. b y integration, e.g. v=a dt    v=u+atv = \int{a}~dt \implies v = u + at,
  2. by using and interpreting appropriate graphs, e.g. velocity against time,
  3. by substitution of one (given) formula into another (given) formula, e.g. substituting v=u+atv = u + at into s=12(u+v)ts = \frac{1}{2}(u+v)t to obtain s=ut+12at2s = ut + \frac{1}{2}at^2.*
#3.02e

Be able to extend the constant acceleration formulae to motion in two dimensions using vectors: v=u+at\bold{v} = \bold{u} + \bold{a}ts=ut+12at2\bold{s} = \bold{u}t + \frac{1}{2}\bold{a}t^2s=12(u+v)t\bold{s} = \frac{1}{2}(\bold{u}+\bold{v})ts=vt12at2\bold{s} = \bold{v}t - \frac{1}{2}\bold{a}t^2

*Questions set involving vectors may involve either column vector notation, e.g. u=(u1u2)\bold{u} = \begin{pmatrix} u_1 \\ u_2 \end{pmatrix} or i\bold{i}, j\bold{j} notation, e.g. u=u1i+u2j\bold{u} = u_1\bold{i} + u_2\bold{j}.

The formula vv=uu+2as\bold{v}⋅\bold{v} = \bold{u}⋅\bold{u} + 2\bold{a}⋅\bold{s} is excluded.*

#3.02f

Be able to use differentiation and integration with respect to time in one dimension to solve simple problems concerning the displacement, velocity and acceleration of a particle:

v=dsdtv = \dfrac{ds}{dt}

a=dvdt=d2sdt2a = \dfrac{dv}{dt} = \dfrac{d^2s}{dt^2}

s=v dts = \displaystyle\int{v}~dt and v=a dtv = \displaystyle\int{a}~dt

#3.02g

Be able to extend the application of differentiation and integration to two dimensions using vectors:

x=f(t)i+g(t)j\bold{x} = f(t)\bold{i} + g(t)\bold{j}

v=dxdt=x˙=f(t)i+g(t)j\bold{v} = \dfrac{d\bold{x}}{dt} = \dot{\bold{x}} = f'(t)\bold{i} + g'(t)\bold{j}

a=dvdt=v˙=f(t)i+g(t)j\bold{a} = \dfrac{d\bold{v}}{dt} = \dot{\bold{v}} = f''(t)\bold{i} + g''(t)\bold{j}

x=v dt\bold{x} = \displaystyle\int{\bold{v}}~dt and v=a dt\bold{v} = \displaystyle\int{\bold{a}}~dt

Questions set may involve either column vector or i\bold{i}, j\bold{j} notation.

#3.02h

Be able to model motion under gravity in a vertical plane using vectors where a=(0g)\bold{a} = \begin{pmatrix} 0 \\ -g \end{pmatrix} or a=gj\bold{a} = -g\bold{j}.

#3.02i

Be able to model the motion of a projectile as a particle moving with constant acceleration and understand the limitation of this model.

*Includes being able to:

  1. Use horizontal and vertical equations of motion to solve problems on the motion of projectiles.
  2. Find the magnitude and direction of the velocity at a given time or position.
  3. Find the range on a horizontal plane and the greatest height achieved.
  4. Derive and use the cartesian equation of the trajectory of a projectile.

Projectiles on an inclined plane and problems with resistive forces are excluded.*

3.01
Quantities and units in mechanics
3.03
Forces and Newton's laws