solve problems using equations that represent uniformly accelerated motion in a straight line, including the motion of bodies falling in a uniform gravitational field without air resistance

Equations of Motion

In this unit we consider motion in a straight line with constant (uniform) acceleration. The equations derived below are the tools used to solve quantitative problems, including the free‑fall of bodies in a uniform gravitational field when air resistance is neglected.

Key Definitions

• Displacement (\$s\$) – change in position along a straight line (m).
• Initial velocity (\$u\$) – velocity at the start of the time interval (m s⁻¹).
• Final velocity (\$v\$) – velocity at the end of the time interval (m s⁻¹).
• Acceleration (\$a\$) – rate of change of velocity (m s⁻²). For free fall \$a = g = 9.81\ \text{m s}^{-2}\$ downwards.
• Time (\$t\$) – elapsed time (s).

Core Equations for Uniformly Accelerated Motion

The four equations below are inter‑related; any three can be combined to eliminate a variable.

1. \$v = u + a t\$
2. \$s = u t + \tfrac{1}{2} a t^{2}\$
3. \$v^{2} = u^{2} + 2 a s\$
4. \$s = \tfrac{1}{2}\,(u+v)\,t\$

Sign Conventions

Choose a positive direction (usually upwards). Then:

• Displacements and velocities in the chosen direction are positive; opposite direction is negative.
• For free fall, if upwards is positive, \$a = -g\$; if downwards is positive, \$a = +g\$.

Derivation of the Equations (Brief)

Starting from the definition of acceleration:

\$a = \frac{dv}{dt}\$

Integrating with respect to time gives the first equation \$v = u + a t\$. Substituting \$v\$ into the definition of displacement \$s = \int v\,dt\$ yields the second equation, and eliminating \$t\$ between the first two gives the third. The fourth equation follows from the definition of average velocity for constant acceleration.

Application to Free Fall

When air resistance is negligible, the only acceleration is due to gravity \$g = 9.81\ \text{m s}^{-2}\$. The equations become:

• \$v = u \pm g t\$
• \$s = u t \pm \tfrac{1}{2} g t^{2}\$
• \$v^{2} = u^{2} \pm 2 g s\$

Use “+” when the chosen positive direction is downwards, “–” when it is upwards.

Worked Example 1 – Dropping a Stone

A stone is released from rest at a height of \$20\ \text{m}\$ above the ground. Find the time taken to reach the ground and its impact speed. Take \$g = 9.81\ \text{m s}^{-2}\$ and choose upwards as positive.

1. Identify known quantities: \$u = 0\$, \$s = -20\ \text{m}\$ (downwards), \$a = -g = -9.81\ \text{m s}^{-2}\$.
2. Use \$s = u t + \tfrac{1}{2} a t^{2}\$:

   \$-20 = 0 \times t + \tfrac{1}{2}(-9.81) t^{2}\$

   \$t^{2} = \frac{2 \times 20}{9.81} \approx 4.08\$

   \$t \approx 2.02\ \text{s}\$

3. Find final speed with \$v = u + a t\$:

   \$v = 0 + (-9.81)(2.02) \approx -19.8\ \text{m s}^{-1}\$

   The magnitude of the impact speed is \$19.8\ \text{m s}^{-1}\$ downwards.

Worked Example 2 – Throwing a Ball Upwards

A ball is thrown vertically upwards with an initial speed of \$15\ \text{m s}^{-1}\$. Determine (a) the maximum height above the launch point, and (b) the total time the ball is in the air.

1. At the highest point \$v = 0\$. Use \$v^{2} = u^{2} + 2 a s\$ with \$a = -g\$:

   \$0 = (15)^{2} - 2(9.81)s\$

   \$s = \frac{15^{2}}{2 \times 9.81} \approx 11.5\ \text{m}\$

2. Time to reach the top from \$v = u + a t\$:

   \$0 = 15 - 9.81 t\$

   \$t_{\text{up}} = \frac{15}{9.81} \approx 1.53\ \text{s}\$

3. Total time is double the ascent time (symmetry of motion):

   \$t_{\text{total}} = 2 \times 1.53 \approx 3.06\ \text{s}\$

Common Mistakes and How to Avoid Them

• Incorrect sign convention: Write down the chosen positive direction before substituting numbers.
• Mixing units: Keep all quantities in SI units (m, s, m s⁻¹, m s⁻²).
• Using the wrong equation: Identify which variables are known and which are required; select the equation that contains only those variables.
• Forgetting that \$s\$ is displacement, not distance travelled: Displacement can be negative; distance is always positive.

Summary Table of Equations

Practice Questions

1. A car accelerates uniformly from \$5\ \text{m s}^{-1}\$ to \$25\ \text{m s}^{-1}\$ in \$8\ \text{s}\$. Calculate the acceleration and the distance covered.
2. A stone is thrown downwards from a cliff \$30\ \text{m}\$ high with an initial speed of \$4\ \text{m s}^{-1}\$. Find the time to reach the base of the cliff and the impact speed.
3. A ball is projected vertically upwards with speed \$20\ \text{m s}^{-1}\$. Determine the height at which its speed is \$10\ \text{m s}^{-1}\$ on the way up and on the way down.

Further Reading

For deeper insight into the derivation of the equations from calculus, consult the A‑Level textbook sections on kinematics and the mathematical appendix on integration of constant acceleration.