Physics — One-Dimensional Kinematics

Physics — One-Dimensional Kinematics

Kinematics is the branch of mechanics that describes how objects move — without considering the forces that caused the motion. Using a few equations of motion we can predict the velocity, position and time of an object moving with constant acceleration. On this page we will introduce displacement, velocity and acceleration, study the three equations of motion, practise free fall, and learn to convert between km/h and m/s.

Background and Basic Definitions

Fundamental quantities:

  • Displacement \( \Delta x = x_2 - x_1 \) — the change in position (a vector quantity; can be negative).
  • Velocity \( v = \frac{\Delta x}{\Delta t} \) — rate of change of position.
  • Acceleration \( a = \frac{\Delta v}{\Delta t} \) — rate of change of velocity.

Three equations of motion for constant acceleration:

\[ v = v_0 + a t \]\[ x = v_0 t + \tfrac{1}{2} a t^2 \]\[ v^2 = v_0^2 + 2 a \Delta x \]

The third equation is especially useful when no time is given.

Free fall is motion under gravitational acceleration \( g \approx 10 \, \text{m/s}^2 \) (downward). In free-fall problems set \( a = g \); for an upward throw set \( a = -g \) (decelerating the object).

Unit conversion: from km/h to m/s divide by \(3.6\); from m/s to km/h multiply by \(3.6\). For example \( 36 \, \text{km/h} = \frac{36}{3.6} = 10 \, \text{m/s} \).

Graphs: on a velocity-time (\(v\text{-}t\)) graph, the slope is the acceleration and the area under the curve is the displacement.

Solution Steps

  1. Step 1 — List the known quantities (\(v_0\), \(v\), \(a\), \(t\), \(\Delta x\)) and convert all units to m/s and m.
  2. Step 2 — Choose a positive direction (e.g. upward) and assign signs: braking acceleration and gravity downward are usually negative.
  3. Step 3 — Select the equation in which the only unknown is the quantity being sought; if no time is given, use \( v^2 = v_0^2 + 2a\Delta x \).
  4. Step 4 — Substitute and solve algebraically for the unknown.
  5. Step 5 — For graph problems: slope of \(v\text{-}t\) = acceleration; area under \(v\text{-}t\) = displacement.
  6. Step 6 — Check reasonableness: the sign of velocity matches the chosen direction, and the magnitude is physically plausible.

Worked Examples

Example 1: Unit Conversion and Acceleration

Problem: A car accelerates from rest (\(v_0 = 0\)) to \(108\) km/h in \(5\) seconds. What is its acceleration in \( \text{m/s}^2 \)?

Solution:

  1. Convert the final speed: \( v = \frac{108}{3.6} = 30 \, \text{m/s} \).
  2. Use \( v = v_0 + a t \) and isolate \(a\): \( a = \frac{v - v_0}{t} = \frac{30 - 0}{5} \).
  3. Compute: \( \frac{30}{5} = 6 \).

Answer: The acceleration is \( 6 \, \text{m/s}^2 \).

Example 2: Final Velocity Under Constant Acceleration

Problem: A car travels at \( v_0 = 10 \, \text{m/s} \) and accelerates at \( a = 2 \, \text{m/s}^2 \) for \( t = 5 \) seconds. What is its final velocity?

Solution:

  1. The appropriate equation is \( v = v_0 + a t \).
  2. Substitute: \( v = 10 + 2 \times 5 = 10 + 10 \).
  3. Compute: \( v = 20 \, \text{m/s} \).

Answer: The final velocity is \( 20 \, \text{m/s} \).

Example 3: Braking Distance

Problem: A car travelling at \( v_0 = 30 \, \text{m/s} \) brakes with acceleration \( a = -5 \, \text{m/s}^2 \) until it stops completely. What distance does it travel before stopping?

Solution:

  1. No time is given, so use \( v^2 = v_0^2 + 2 a \Delta x \) with \( v = 0 \) (stopped).
  2. Substitute: \( 0 = 30^2 + 2 \times (-5) \times \Delta x = 900 - 10 \Delta x \).
  3. Isolate: \( 10 \Delta x = 900 \), so \( \Delta x = 90 \, \text{m} \).

Answer: The braking distance is \( 90 \) metres.

Example 4: Free-Fall Time

Problem: A stone is dropped from rest (\( v_0 = 0 \)) from a height of \( h = 80 \) metres. How long does it take to reach the ground? (\( g = 10 \, \text{m/s}^2 \))

Solution:

  1. For free fall from rest: \( h = \tfrac{1}{2} g t^2 \).
  2. Substitute: \( 80 = \tfrac{1}{2} \times 10 \times t^2 = 5 t^2 \).
  3. Isolate: \( t^2 = \frac{80}{5} = 16 \), so \( t = 4 \) seconds.

Answer: The fall takes \( 4 \) seconds.

Example 5: Upward Throw — Impact Speed

Problem: From a rooftop \(20\) metres high, a ball is thrown upward at \( 10 \, \text{m/s} \). At what speed does it hit the ground? (\( g = 10 \, \text{m/s}^2 \))

Solution:

  1. Choose downward as positive. Then \( v_0 = -10 \, \text{m/s} \) (thrown upward), \( a = 10 \, \text{m/s}^2 \), and displacement to ground \( \Delta x = 20 \) m.
  2. Use \( v^2 = v_0^2 + 2 a \Delta x \): \( v^2 = (-10)^2 + 2 \times 10 \times 20 = 100 + 400 = 500 \).
  3. Take the square root: \( v = \sqrt{500} \approx 22.36 \, \text{m/s} \).
  4. Intuition: the ball rises, returns to roof level at \(10\) m/s downward, then falls another \(20\) m.

Answer: The impact speed is \( \sqrt{500} \approx 22.4 \, \text{m/s} \).

Common Mistakes

✗ Common mistake: Substituting speed in km/h directly into the equations of motion alongside \(a\) in \( \text{m/s}^2 \).

✓ The correct way: The equations require consistent units. Convert km/h to m/s first (divide by \(3.6\)) before substituting.

✗ Common mistake: Ignoring the sign of acceleration — entering braking or gravitational acceleration as positive when it should be negative.

✓ The correct way: Acceleration opposing the direction of motion (braking) or acting downward when upward is positive is negative. Fix the positive direction at the start of the solution and stick to it.

✗ Common mistake: Thinking that doubling the speed doubles the braking distance.

✓ The correct way: From \( v^2 = v_0^2 + 2a\Delta x \) it follows that \( \Delta x \propto v_0^2 \). Doubling the speed quadruples the braking distance, not doubles it.

Practice Tips

  • Tip — Remember the conversion: km/h \(\to\) m/s divide by \(3.6\); m/s \(\to\) km/h multiply by \(3.6\).
  • Tip — When time is not given in the problem, the correct equation is almost always \( v^2 = v_0^2 + 2a\Delta x \).
  • Tip — On a \(v\text{-}t\) graph: slope = acceleration; area under the curve = displacement. A horizontal graph means constant velocity (zero acceleration).
  • Tip — At the highest point of an upward throw the instantaneous velocity is zero; this is a convenient starting point for calculating height or time.

Summary and Key Formulas

  • Definitions: \( v = \frac{\Delta x}{\Delta t} \), \( a = \frac{\Delta v}{\Delta t} \).
  • Equations of motion: \( v = v_0 + at \); \( x = v_0 t + \tfrac{1}{2} a t^2 \); \( v^2 = v_0^2 + 2 a \Delta x \).
  • Free fall: \( a = g \approx 10 \, \text{m/s}^2 \).
  • Conversion: km/h \(\div 3.6 =\) m/s.
  • \(v\text{-}t\) graph: slope = acceleration; area = displacement.
  • Braking distance is proportional to \( v_0^2 \).