2.1.6Physical quantities and kinematics
Deriving Kinematic Equations
5 min read 10 parts Lesson 15 of 22 1 quiz
Lesson summary
2.1.6 Deriving Kinematic Equations — AS PHYSICS 9702 notes: Derive the four equations of uniformly accelerated motion from the definition of acceleration and a velocity-time graph.Four equations describe any object moving with constant acceleration in a straight line.They relate …
- They relate five variables
- Each equation leaves out one variable
- What the symbols mean
- They are valid only for constant acceleration
- They are vector equations
- Multiplying by t gives at = v − u
Section 1 of 10The equations of uniform acceleration
The equations of uniform acceleration
Four equations describe any object moving with constant acceleration in a straight line.
They relate five variables
- They relate five variables displacement s, initial velocity u, final velocity v, acceleration a and time t.
Each equation leaves out one variable
- Each equation leaves out one variable v = u + at (no s), s = ½(u + v)t (no a), s = ut + ½at2 (no v), v2 = u2 + 2as (no t).What the symbols mean: v = final velocity (m s−1) · u = initial velocity (m s−1) · a = acceleration (m s−2) · t = time (s)
They are valid only for constant acceleration
- They are valid only for constant acceleration along a straight line.
They are vector equations
- They are vector equations so choose a positive direction and use negative values for opposite quantities.
Section 2 of 10Derivation of v = u + at
Derivation of v = u + at
By definition a = (v − u) ÷ t.
Multiplying by t gives at
Multiplying by t gives at v − u
- and so v = u + at.
On a v-t graph this is a straight line
- On a v-t graph this is a straight line with intercept u and gradient a, so the height at time t is u + at.
Section 3 of 10Derivation of s = ½(u + v)t
1
The displacement is the area under the v-t graph.
2For uniform acceleration the graph is a straight line from u to v
- For uniform acceleration the graph is a straight line from u to v it forms a trapezium with parallel sides u and v and width t.
3
The area is ½(u + v)t.
4Equivalently the average velocity is halfway between u and v
- Equivalently the average velocity is halfway between u and v so displacement = average velocity × time.
Section 4 of 10Derivation of s = ut + ½at²
Derivation of s = ut + ½at²
Substitute v = u + at into s = ½(u + v)t.
This gives s = ½(2u + at)t = ut + ½at2.
On the graph the area splits into a rectangle ut
- On the graph the area splits into a rectangle ut and a triangle ½(at)(t).
Section 5 of 10Derivation of v² = u² + 2as
1
From v = u + at, t = (v − u) ÷ a.
2Substitute into s
½(u + v)t
- s = ½(u + v)(v − u) ÷ a = (v2 − u2) ÷ 2a.
3
Rearranging gives v2 = u2 + 2as.
4The equation links velocity, acceleration and distance
- The equation links velocity, acceleration and distance without time.
Section 6 of 10Special cases and checks
1Starting from rest, u = 0
- Starting from rest, u = 0 gives s = ½at2 and v2 = 2as.
2Zero acceleration, a = 0
- Zero acceleration, a = 0 all the equations reduce to v = u and s = vt.
3Check by units
- Check by units every term in s = ut + ½at2 has the dimension of length.
4Only s
Only s ut + ½at² and v² = u² + 2as are given on the data sheet
- the other two must be remembered.
Section 7 of 10Links to later topics
1v2
v2 u2 + 2as multiplied by m ÷ 2
- becomes ½mv2 − ½mu2 = Fs, the work-energy relation, because F = ma.
2v
v u + at multiplied by m
- gives mv − mu = Ft, the impulse-momentum relation.
3Kinematics and energy
- Kinematics and energy are two views of one idea.
4In a derivation state the starting point
- In a derivation state the starting point either the definition a = (v − u) ÷ t or the area under the v-t graph, and that a is constant.
Section 8 of 10More from the specification
More from the specification
Timing vehicle speed using light gates
- Timing vehicle speed using light gates interrupting infrared beams of digital light gates gives accurate instantaneous speed without human reflex latency.
Velocity from Displacement-Time Graphs
- On a displacement-time graph, the gradient or slope of the curve represents the instantaneous velocity of the object,
- whereas a horizontal line indicates zero velocity.
Acceleration and Velocity-Time Graphs
- On a velocity-time graph, the gradient equals acceleration,
- and the area enclosed beneath the graph represents the displacement change over the time interval taken.
Experimental Determination of g
- Measuring the time taken for a falling object such as a steel ball-bearing to traverse known heights allows acceleration of free fall g to be determined using kinematic equations.
Velocity from Displacement-Time Graphs
- On a displacement-time graph, the gradient or slope of the curve represents the instantaneous velocity of the object,
- whereas a horizontal line indicates zero velocity.
Acceleration and Velocity-Time Graphs
- On a velocity-time graph, the gradient equals acceleration,
- and the area enclosed beneath the graph represents the displacement change over the time interval taken.
Experimental Determination of g
- Measuring the time taken for a falling object such as a steel ball-bearing to traverse known heights allows acceleration of free fall g to be determined using kinematic equations.
Displacement-time graph
- A graph of displacement against time whose gradient is velocity.
Acceleration-time graph
- A graph of acceleration against time whose area is the change in velocity.
Light gate
- A sensor that starts or stops a timer when a beam is interrupted.
Scalar quantity
- A quantity with magnitude only, such as potential
Projectile
- Object moving under gravity alone in two dimensions.
Electromagnet
- Magnet made by current through a coil around a magnetic core.
Section 9 of 10Key terms
| Term | Meaning |
|---|---|
| Uniform acceleration | Acceleration that has a constant size and direction. |
| Initial velocity u | The velocity at the start of the time interval. |
| Final velocity v | The velocity at the end of the time interval. |
| Displacement s | The change in position in a stated direction. |
| Average velocity | For uniform acceleration, ½(u + v). |
| Kinematic equations | The four equations of motion for constant acceleration. |
| Trapezium | The shape under a straight v-t line, with area equal to displacement. |
| Data sheet | The list of equations given in the examination. |
Section 10 of 10Exam technique
1
Applying the equations when acceleration is not constant, for example across a change of braking force.
2
Mixing up u and v, particularly for an object that has been decelerating.
3
Dropping signs — a quantity opposite to the chosen positive direction must be negative.
4
Quoting the derivation without stating that a is constant.