Uniform and Variable Acceleration: MDCAT Physics notes

Uniform and variable acceleration

Uniform acceleration: the velocity changes by equal amounts in equal intervals of time, however small the intervals. Its velocity–time graph is a straight line of constant slope.

Variable acceleration: the velocity changes by unequal amounts in equal intervals of time. The v–t graph is a curve; its slope differs from point to point, so the acceleration is changing.

• Average acceleration: $a_{av} = \Delta v/\Delta t$ over a finite interval.
• Instantaneous acceleration: the limit of $\Delta v/\Delta t$ as $\Delta t \to 0$, i.e. the slope of the tangent to the v–t graph at that instant.
• For uniform acceleration, average and instantaneous accelerations are equal.

Direction of acceleration

Acceleration is a vector. Its direction is the direction of the change in velocity, not necessarily the direction of motion.

Rule: when a body slows down, its acceleration points opposite to its velocity. This is called deceleration or retardation; in calculations it enters as a negative value.

Key formulas (uniform acceleration only)

$$v_f = v_i + at$$

$$S = v_i t + \tfrac12 a t^2$$

$$2aS = v_f^2 - v_i^2$$

$$S = \frac{v_i + v_f}{2}\,t$$

For free fall near the Earth, $a = g = 9.8\ \mathrm{m\,s^{-2}}$ downward. A body dropped from rest ($v_i = 0$) falls

$$S = \tfrac12 g t^2 = 4.9\,t^2 \ \text{metres}$$

and has speed $v = gt$ or $v = \sqrt{2gh}$ after falling a height $h$.

Worked examples

1. Starting from rest. A bike starts from rest with uniform acceleration $2\ \mathrm{m\,s^{-2}}$. After 6 s: $v_f = 0 + 2\times6 = 12\ \mathrm{m\,s^{-1}}$; distance $= \tfrac12\times2\times6^2 = 36$ m.

2. Braking. A car moving at $20\ \mathrm{m\,s^{-1}}$ must stop within 25 m. Using $v_f^2 = v_i^2 + 2aS$: $0 = 400 + 2a(25)$, so $a = -8\ \mathrm{m\,s^{-2}}$. The magnitude of deceleration is $8\ \mathrm{m\,s^{-2}}$.

3. Partway through a fall. A ball is released from rest 10 m above the ground. Its speed when it is 5 m above the ground depends only on the height fallen so far, 5 m: $v = \sqrt{2\times9.8\times5} = \sqrt{98} \approx 9.9\ \mathrm{m\,s^{-1}}$. The mass does not matter.

4. Free-fall distance. In 3 s a dropped stone falls $4.9\times3^2 = 44.1$ m.

Common MDCAT traps

• "Unequal changes in velocity in equal times" means variable acceleration, not uniform.
• For a body slowing along −x, acceleration is along +x.
• In free-fall problems use the height actually fallen, not the height above the ground.
• $S = 4.9t^2$ comes from $\tfrac12 g$; do not use $9.8t^2$.
• A changing slope on a v–t graph means changing acceleration, not uniform velocity.

Quick revision

• Uniform acceleration: straight v–t line.
• Variable acceleration: curved v–t line.
• Deceleration: acceleration opposite to velocity.
• $g = 9.8\ \mathrm{m\,s^{-2}}$; dropped body falls $4.9t^2$ m.
• Equations of motion apply only for uniform acceleration.