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 📐

1. Uniformly Accelerated Motion

When a body moves in a straight line and its speed changes at a constant rate, we call that uniform acceleration. Think of a car that starts from a stoplight and speeds up steadily – that’s uniform acceleration. The key equations are:

  • \$s = ut + \frac12 at^2\$ – distance travelled

  • \$v = u + at\$ – final velocity

  • \$v^2 = u^2 + 2as\$ – relates speed and distance without time

  • \$s = \frac12 (u+v)t\$ – average speed times time

Analogy: Imagine a skateboarder who pushes off the ground and keeps pushing at the same rate. The push is the constant acceleration, and the distance they cover follows the equations above.

Example Problem

A car starts from rest (\$u=0\$) and accelerates at \$2\,\text{m/s}^2\$. Find the distance it travels in \$5\,\text{s}\$.

  1. Insert values into \$s = ut + \tfrac12 at^2\$:

  2. \$s = 0 \times 5 + \tfrac12 (2)(5)^2 = 0 + 25 = 25\,\text{m}\$

  3. Answer: \$25\,\text{m}\$.

Exam Tip: Always check the units – acceleration is in m/s², time in s, distance in m. A missing factor of ½ can lead to a 2× error! 🚀

2. Falling Bodies in a Uniform Gravitational Field

When a body falls near Earth’s surface and we ignore air resistance, the only force is gravity. The acceleration is constant: \$g = 9.81\,\text{m/s}^2\$ downward. We can treat this as a special case of uniform acceleration with \$a = g\$.

Key equations (take upward as positive):

  • \$s = ut + \frac12 gt^2\$

  • \$v = u + gt\$

  • \$v^2 = u^2 + 2gs\$

Example: A ball is dropped from a height of \$20\,\text{m}\$ (so \$u=0\$, \$s=-20\,\text{m}\$). Find the time to hit the ground and its impact speed.

  1. Time: \$s = \tfrac12 gt^2 \Rightarrow -20 = \tfrac12 (9.81)t^2\$

  2. \$t^2 = \frac{40}{9.81} \approx 4.08 \Rightarrow t \approx 2.02\,\text{s}\$

  3. Speed: \$v = gt = 9.81 \times 2.02 \approx 19.8\,\text{m/s}\$ downward.

Exam Tip: Remember to keep the sign of \$s\$ consistent with your chosen direction. If you take downward as positive, \$s=+20\,\text{m}\$ and \$g=+9.81\,\text{m/s}^2\$. This avoids confusion in the equations. 📏

3. Common Mistakes & Quick Checks

  • Mixing up initial and final velocities – label them \$u\$ and \$v\$ clearly.

  • Forgetting the ½ factor in \$s = ut + \tfrac12 at^2\$.

  • Using \$g = 9.8\$ instead of \$9.81\$ can give a small error; use the value given in the question.

  • Sign errors: choose a direction (upward or downward) and stick with it throughout.

Quick Check List for Exams:

  1. Identify knowns and unknowns.

  2. Select the correct equation (avoid using \$v = u + at\$ if time is unknown).

  3. Plug in numbers carefully – double‑check each value.

  4. Verify units and sign.

  5. Show all steps; partial credit is often awarded for clear reasoning.

4. Practice Table

ScenarioKnownsUnknownEquation Used
Car accelerates from rest\$u=0\$, \$a=3\,\text{m/s}^2\$, \$t=4\,\text{s}\$\$s\$\$s = ut + \tfrac12 at^2\$
Ball dropped from 15 m\$u=0\$, \$s=-15\,\text{m}\$, \$g=9.81\,\text{m/s}^2\$\$t\$\$s = \tfrac12 gt^2\$
Projectile launched upward\$u=20\,\text{m/s}\$, \$a=-9.81\,\text{m/s}^2\$, \$t=3\,\text{s}\$\$v\$\$v = u + at\$

Final Exam Tip: When you’re in a hurry, remember the “quick‑look” method: pick the equation that contains the unknown you need and only use the variables you’re given. If you’re unsure, write down all equations first and then decide. Good luck! 🍀