# Displacement and Velocity-Time Graphs ## Displacement-Time Graphs - Displacement-time graphs illustrate how an object's position changes as time progresses - They reveal whether an object travels forwards (positive displacement values) or backwards (negative displacement values)

velocity = gradient of a displacement-time graph

### Interpreting the Gradient - A steeper gradient indicates greater velocity - A negative gradient means negative velocity (object moving backwards/in the opposite direction) - A horizontal line (zero gradient) shows the object is stationary - A straight diagonal line represents constant velocity - A curved line indicates changing velocity (acceleration or deceleration) **Example displacement-time graphs showing different motion types:**

### Finding Instantaneous Velocity - For objects with changing velocity, draw a tangent line at the specific time point of interest - Calculate the gradient of that tangent to find the instantaneous velocity - Remember: instantaneous velocity at a point differs from average velocity over an interval ### Worked Example: Converting Motion Graphs Consider a train that travels at steady speed, then applies brakes gradually until it stops at a station. **Given velocity-time graph:** - Horizontal line from $t = 0$ to $t = 8$ s - Straight diagonal line (downward) from $t = 8$ s to $t = 15$ s

**Corresponding displacement-time graph:** 1. When velocity is constant (horizontal on v-t graph): - Displacement-time shows a straight diagonal line with constant gradient 2. When velocity decreases uniformly (diagonal on v-t graph): - Displacement-time shows a curve with decreasing gradient - The gradient becomes flatter as the train slows down - Eventually reaches horizontal when velocity becomes zero

## Velocity-Time Graphs - Velocity-time graphs display both the speed and direction of motion over time - Unlike speed-time graphs (always positive), velocity-time graphs can show negative values indicating reversed direction

displacement = area under a velocity-time graph

### Understanding Acceleration - Acceleration represents any change in velocity over time - Since velocity is a vector, changes in either speed or direction constitute acceleration - Objects that slow down are commonly described as 'decelerating'

acceleration = gradient of a velocity-time graph

### Interpreting Features | Feature | Meaning | |---------|---------| | Straight diagonal line | Uniform (constant) acceleration | | Positive gradient | Increasing velocity in positive direction | | Negative gradient | Increasing velocity in negative direction (or deceleration if already positive) | | Curved line | Non-uniform acceleration | | Horizontal line (zero gradient) | Constant velocity (zero acceleration) | **Example velocity-time graph showing uniform acceleration:**

### Calculating Displacement - The displacement equals the total area between the graph line and the time axis - For complex shapes, break the area into triangles, rectangles, and trapeziums - Areas below the time axis (negative velocity) represent negative displacement ## Motion of a Bouncing Ball Consider a ball thrown upwards that bounces on the ground repeatedly: ### Choosing Direction Convention - Define positive direction as upwards - Negative direction is therefore downwards - Maintain consistency throughout analysis ### Key Points in the Motion **At the highest point:** - Ball reaches maximum displacement - Velocity momentarily equals zero - Velocity transitions from positive to negative (changes direction) - Acceleration remains constant at $g = 9.81 \text{ m s}^{-2}$ downwards throughout **At the lowest point (ground level):** - Ball reaches minimum displacement (zero height) - Velocity instantaneously switches from negative to positive - The speed magnitude stays the same, but direction reverses - Sudden direction change creates a momentary large acceleration ### Velocity-Time Graph Characteristics - Ball travelling upwards: positive velocity decreasing towards zero - At peak: velocity crosses zero, graph intersects time axis - Ball falling downwards: negative velocity increasing in magnitude - At bounce: sharp V-shape as velocity instantly reverses - Assuming no air resistance, the pattern repeats symmetrically **Velocity-time graph for a bouncing ball (with some energy loss):**

### Acceleration-Time Graph Characteristics - Horizontal line at $a = -9.81 \text{ m s}^{-2}$ throughout flight - Brief spikes upwards at each bounce (ground exerts large upward force) - Between bounces, acceleration constant (gravity only) **Acceleration-time graph for a bouncing ball:**

Velocity changes sign when direction changes - don't confuse this with speed, which remains positive

## Summary: Gradients and Areas Understanding what different graph features represent: | Graph Type | Gradient Represents | Area Under Graph Represents | |------------|-------------------|---------------------------| | Displacement-Time | Velocity | No physical meaning | | Velocity-Time | Acceleration | Displacement | | Acceleration-Time | No physical meaning | Change in velocity | ### Practical Tips - Always check axis labels carefully before analysing any motion graph - For curved sections, use tangents to find instantaneous rates - Break complex areas into simple geometric shapes (triangles, rectangles, trapeziums) - Remember to include appropriate units in all calculations ## Worked Example: Finding Displacement from Area A motorcycle accelerates uniformly from rest. Its velocity-time graph shows: - Straight line from $(0, 0)$ to $(12 \text{ s}, 25 \text{ m s}^{-1})$

Calculate the displacement after 12 s. **Answer:** **Step 1:** Recognise the graph forms a triangle - Base = time = 12 s - Height = final velocity = 25 m s$^{-1}$ **Step 2:** Calculate area of triangle $$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$$ $$\text{Displacement} = \frac{1}{2} \times 12 \times 25$$ $$\text{Displacement} = 150 \text{ m}$$ **Alternative method:** Average velocity $$\text{Average velocity} = \frac{0 + 25}{2} = 12.5 \text{ m s}^{-1}$$ $$\text{Displacement} = \text{average velocity} \times \text{time} = 12.5 \times 12 = 150 \text{ m}$$ ### Example: Non-Linear Graph A cyclist's velocity changes non-uniformly. To estimate displacement: 1. Divide the area under the curve into strips (trapeziums or rectangles) 2. Calculate the area of each strip separately 3. Sum all areas to find total displacement 4. Smaller strips provide more accurate estimates