derive, from the definitions of velocity and acceleration, equations that represent uniformly accelerated motion in a straight line

Uniformly Accelerated Motion – Derivation and Applications (Cambridge IGCSE/A‑Level)

1. Fundamental Definitions

• Distance \(d\) (m) – scalar, total length of the path travelled.
• Displacement \(s\) (or \(x\)) (m) – vector, change in position along a straight line (positive in the chosen direction).
• Speed \(|v|\) (m s^‑1) – scalar magnitude of velocity.
• Velocity \(v\) (m s^‑1) – vector rate of change of displacement

  \[

  v=\frac{ds}{dt}

  \]

• Acceleration \(a\) (m s^‑2) – vector rate of change of velocity

  \[

  a=\frac{dv}{dt}

  \]

• Initial velocity \(u\) – velocity at \(t=0\).
• Final velocity \(v\) – velocity at a later instant \(t\).

1.1 Sign‑convention reminder

Choose a single direction as positive (e.g. rightward or upward). All vectors are given a sign according to this choice. A negative acceleration means the acceleration vector points opposite to the positive direction.

2. Uniform (Constant) Acceleration

When \(a\) is constant the definitions can be integrated directly, giving the four standard kinematic equations.

2.1 Derivation of \(v = u + at\)

1. Start from the definition \(a = \dfrac{dv}{dt}\).
2. Separate variables ( \(a\) is constant): \(dv = a\,dt\).
3. Integrate between the known start \((t=0,\;v=u)\) and a general instant \((t,\;v)\):

   \[

   \int{u}^{v} dv = a\int{0}^{t} dt

   \]

4. Result: \(v-u = at\).
5. Re‑arrange:

   \[

   \boxed{v = u + at}

   \]

2.2 Derivation of \(s = ut + \tfrac12 a t^{2}\)

1. Use the definition \(v = \dfrac{ds}{dt}\).
2. Substitute the expression for \(v\) from 2.1:

   \[

   \frac{ds}{dt}=u+at

   \]

3. Separate variables: \(ds = (u+at)\,dt\).
4. Integrate from \((t=0,\;s=0)\) to \((t,\;s)\):

   \[

   \int{0}^{s} ds = \int{0}^{t} (u+at)\,dt

   \]

5. Perform the integration:

   \[

   s = ut + \frac12 a t^{2}

   \]

6. Thus

   \[

   \boxed{s = ut + \tfrac12 a t^{2}}

   \]

2.3 Derivation of \(v^{2}=u^{2}+2as\)

1. From \(v = u + at\) solve for the time:

   \[

   t = \frac{v-u}{a}

   \]

2. Insert this expression for \(t\) into the displacement equation of 2.2:

   \[

   s = u\!\left(\frac{v-u}{a}\right) + \frac12 a\!\left(\frac{v-u}{a}\right)^{2}

   \]

3. Algebraic simplification (shown step‑by‑step):

   \[

   \begin{aligned}

   s &= \frac{u(v-u)}{a} + \frac12 a\frac{(v-u)^{2}}{a^{2}}\\[4pt]

   &= \frac{u(v-u)}{a} + \frac{(v-u)^{2}}{2a}\\[4pt]

   &= \frac{2u(v-u)+(v-u)^{2}}{2a}\\[4pt]

   &= \frac{(v-u)\bigl[2u+(v-u)\bigr]}{2a}\\[4pt]

   &= \frac{(v-u)(v+u)}{2a}\\[4pt]

   &= \frac{v^{2}-u^{2}}{2a}

   \end{aligned}

   \]

4. Multiply both sides by \(2a\):

   \[

   2as = v^{2} - u^{2}

   \]

5. Re‑arrange:

   \[

   \boxed{v^{2}=u^{2}+2as}

   \]

2.4 Average‑velocity form \(s = \dfrac{u+v}{2}\,t\)

With constant acceleration the velocity varies linearly, so the average velocity over the interval is the arithmetic mean of the end values:

\[

\bar v = \frac{u+v}{2}\qquad\Longrightarrow\qquad s = \bar v\,t = \frac{u+v}{2}\,t

\]

\[

\boxed{s = \dfrac{u+v}{2}\,t}

\]

3. Graphical Interpretation (Syllabus Requirement 2.1)

3.1 Required graphs

3.2 Worked graphical example (extracting displacement)

In a velocity–time diagram the following data are read:

• Initial velocity \(u = 2.0\ \text{m s}^{-1}\) (at \(t=0\)).
• Final velocity \(v = 8.0\ \text{m s}^{-1}\) (at \(t = 4.0\ \text{s}\)).
• The graph is a straight line, so acceleration is constant.

Displacement is the area under the line:

\[

s = \text{area of trapezium} = \frac{(u+v)}{2}\,t

= \frac{(2.0+8.0)}{2}\times4.0

= 20.0\ \text{m}

\]

Using the algebraic equation gives the same result:

\[

s = ut + \tfrac12 a t^{2},\qquad a = \frac{v-u}{t}= \frac{8.0-2.0}{4.0}=1.5\ \text{m s}^{-2}

\]

\[

s = 2.0\times4.0 + \tfrac12(1.5)(4.0)^{2}=8.0+12.0=20.0\ \text{m}

\]

4. Experimental Determination of the Acceleration due to Gravity (\(g\))

4.1 Photogate (or timing‑ball) method – step‑by‑step worksheet

1. Set‑up

   • Mount two photogates vertically on a rigid stand; the separation \(s\) should be measured with a ruler or steel tape (e.g. \(s = 0.500\ \text{m}\) ± 0.001 m).
   • Connect the gates to a timer that records the interval between the first and second interruptions.

2. Procedure

   • Release a small steel ball from rest just above the first gate so that it triggers the start of the timer.
   • The ball then passes the second gate, stopping the timer.
   • Record the time \(t\). Repeat at least 5 times, re‑positioning the ball each trial to minimise systematic error.

3. Data table (example)

   Trial	Time \(t\) (s)
   1	0.319
   2	0.322
   3	0.318
   4	0.321
   5	0.320

   Mean time \(\bar t = 0.320\ \text{s}\) Standard deviation \(\sigma_t \approx 0.0015\ \text{s}\).

4. Calculate \(g\) using the equation for motion from rest:

   \[

   s = \tfrac12 g t^{2}\;\Longrightarrow\; g = \frac{2s}{t^{2}}

   \]

   \[

   g = \frac{2(0.500\ \text{m})}{(0.320\ \text{s})^{2}} = 9.77\ \text{m s}^{-2}

   \]

5. Uncertainty propagation (simple relative‑error method)

   \[

   \frac{\Delta g}{g} = \sqrt{\left(\frac{\Delta s}{s}\right)^{2} + \left(2\frac{\Delta t}{t}\right)^{2}}

   \]

   Using \(\Delta s = 0.001\ \text{m}\) and \(\Delta t = 0.002\ \text{s}\):

   \[

   \frac{\Delta g}{g}= \sqrt{(0.002)^{2} + \bigl(2\times0.0063\bigr)^{2}} \approx 0.013

   \]

   \[

   \Delta g \approx 0.013\times9.77 \approx 0.13\ \text{m s}^{-2}

   \]

   Result: \(g = 9.77 \pm 0.13\ \text{m s}^{-2}\) (agreeing with the accepted \(9.81\ \text{m s}^{-2}\) within uncertainty).

6. Sources of error (to discuss in the report)

   • Air resistance (small for a steel ball).
   • Finite width of the photogate beams – the timer starts/stops when the ball’s leading edge breaks the beam, not its centre.
   • Reaction time is negligible with electronic gates, but systematic mis‑alignment of the gates changes the effective distance \(s\).

5. Motion with Uniform Velocity in One Direction & Uniform Perpendicular Acceleration

5.1 Quantitative projectile example (horizontal launch)

A steel ball is projected horizontally from the top of a 1.20 m high table with an initial speed \(u{x}=4.0\ \text{m s}^{-1}\). Take \(a{y}=g=9.81\ \text{m s}^{-2}\) downwards.

1. Vertical motion (initial vertical speed \(u_{y}=0\)):

   \[

   y = u_{y}t + \tfrac12 g t^{2} = \tfrac12 g t^{2}

   \]

   Set \(y = 1.20\ \text{m}\) and solve for \(t\):

   \[

   t = \sqrt{\frac{2y}{g}} = \sqrt{\frac{2\times1.20}{9.81}} = 0.495\ \text{s}

   \]

2. Horizontal motion (no horizontal acceleration):

   \[

   x = u_{x} t = 4.0 \times 0.495 = 1.98\ \text{m}

   \]

3. Result – the ball lands 2.0 m from the edge of the table after 0.50 s.

5.2 Component‑summary table

6. Summary of the Four Core Kinematic Equations

7. Common Pitfalls (Syllabus Reminder)

• Distance vs. displacement: the equations use displacement \(s\); they do not give the total path length if the motion reverses direction.
• Sign convention: decide on a positive direction at the start and keep it consistent; a negative acceleration means the vector points opposite to that direction.
• Constant‑acceleration assumption: the four equations are valid only when \(a\) remains unchanged throughout the interval.
• Speed vs. velocity: speed is the magnitude \(|v|\); the equations involve the signed velocity \(v\) (including direction).
• Using the wrong graph: remember:

  • Gradient of a \(v\!-\!t\) graph = acceleration.
  • Area under a \(v\!-\!t\) graph = displacement.
  • Gradient of an \(s\!-\!t\) or \(x\!-\!t\) graph = velocity.

8. Practice Questions (Extended Range)

1. If a car starts from rest and accelerates uniformly at \(3.0\ \text{m s}^{-2}\) for \(5.0\ \text{s}\), what is its final speed?
2. A stone is thrown vertically upward with an initial speed of \(20\ \text{m s}^{-1}\). Using \(a = -9.8\ \text{m s}^{-2}\), calculate the maximum height reached.
3. Given \(u = 4\ \text{m s}^{-1}\), \(v = 20\ \text{m s}^{-1}\) and \(a = 4\ \text{m s}^{-2}\), find the displacement \(s\) without using time.
4. A block slides down a smooth (frictionless) incline of length \(L = 2.0\ \text{m}\) that makes an angle \(\theta = 30^{\circ}\) with the horizontal. The block starts from rest at the top. Determine:

   • The acceleration of the block down the plane.
   • The speed of the block when it reaches the bottom.

   (Use \(g = 9.8\ \text{m s}^{-2}\) and the component of gravity parallel to the incline.)

5. Horizontal projectile: a ball is launched from a 1.20 m high table with a horizontal speed of \(4.0\ \text{m s}^{-1}\). Calculate the time of flight and the horizontal range. (Use the method of Section 5.)

These notes give a complete, syllabus‑aligned pathway from the basic definitions of velocity and acceleration to the four kinematic equations, their graphical meanings, experimental determination of \(g\), and the combined‑component situation that underpins projectile motion. They also highlight common errors and provide worked examples to support exam preparation.