Position is the location of an object along a straight line (1D). It is measured relative to an origin and can be positive or negative depending on the direction.
Example: \( x = +5\, m \) means 5 meters to the right of the origin; \( x = -3\, m \) means 3 meters to the left.
Displacement is the change in position of an object. It is a vector quantity and can be positive or negative.
Example: If an object moves from \( x = 2\, m \) to \( x = 7\, m \), then displacement is \( \Delta x = 7 - 2 = +5\, m \).
If it moves from \( x = 7\, m \) to \( x = 3\, m \), then \( \Delta x = 3 - 7 = -4\, m \).
Velocity is the rate of change of position with respect to time. It is a vector quantity with magnitude and direction.
Units: meters per second (m/s)
Example: If displacement is \( +10\, m \) in \( 5\, s \), velocity is \( v = \frac{10\, m}{5\, s} = 2\, m/s \) to the right.
Acceleration is the rate of change of velocity with respect to time. It is a vector quantity.
Units: meters per second squared (m/s²)
Example: If velocity changes from \( 2\, m/s \) to \( 6\, m/s \) in \( 2\, s \), acceleration is \( a = \frac{6 - 2}{2} = 2\, m/s^2 \) in the positive direction.
Below is a position vs time graph showing an object moving with constant acceleration. This graph helps visualize how position changes over time.
What the graph shows:
The motion represented:
This graph shows an object moving with constant acceleration. The position function is:
This represents:
Key features:
Important relationships:
These formulas relate displacement, velocity, acceleration, and time when acceleration is constant. (formulas 4 and 5 aren't given on the equation sheet)
Meaning: Final velocity equals initial velocity plus acceleration times time.
Use: Find the final velocity after a certain time when acceleration is constant.
Example: If a car starts at 5 m/s and accelerates at 2 m/s² for 3 seconds, what is its final velocity?
Solution: \( v = 5 + 2 \times 3 = 11 \, m/s \)
Meaning: Displacement equals initial position plus initial velocity times time plus half acceleration times time squared.
Use: Find position after time \(t\) with known initial velocity and acceleration.
Example: A car starts at 5 m/s at position 10 m and accelerates at 3 m/s² horizontally. Where is it after 4 seconds?
Solution: \( x = 10 + 5 \times 4 + \frac{1}{2} \times 3 \times 4^2 = 10 + 20 + 24 = 54 \, m \)
Meaning: Final velocity squared equals initial velocity squared plus twice acceleration times displacement.
Use: Find final velocity when you know displacement and acceleration but not time.
Example: A car accelerates from rest at 3 m/s² over 100 m. What is its final velocity?
Solution: \( v^2 = 0 + 2 \times 3 \times 100 = 600 \Rightarrow v = \sqrt{600} \approx 24.5 \, m/s \)
Meaning: Displacement equals initial position plus the inital velocity times time IF there is no acceleration.
Use: Calculate displacement when you know initial and final velocities and time.
Example: A bike is going at a constant 6 m/s over 4 seconds. How far does it travel?
Solution: \( x = 0 + 6 \times 4 = 6 \times 4 = 24 \, m \)
Meaning: Similar to formula 2 but used when acceleration acts opposite to velocity direction.
Use: Calculate displacement when object slows down due to acceleration opposite to velocity.
Example: A ball thrown upward at 10 m/s slows down due to gravity -9.8 m/s². What is its position after 1 second?
Solution: \( x = 0 + 10 \times 1 - \frac{1}{2} \times 9.8 \times 1^2 = 10 - 4.9 = 5.1 \, m \)