Physics IGCSE 0625: Comprehensive Guide to Speed-Time and Distance-Time Graphs

Core Principles of Speed-Time Graphs

A speed-time graph serves as a visual representation of how an object's velocity changes over a specific duration. There are two primary rules for interpreting these graphs. First, the gradient or slope of the line indicates the acceleration of the object. A steeper slope represents a greater rate of acceleration, while a horizontal line indicates zero acceleration or constant speed. Second, the total distance travelled by the object is equivalent to the area under the speed-time graph. This area can often be calculated by breaking the shape down into geometric forms such as triangles and rectangles.

Analyzing a Symmetrical Speed-Time Journey

Consider a car travelling between two sets of traffic lights. The motion is depicted on a graph where speed in m/sm/s is plotted on the vertical axis and time in ss is on the horizontal axis. The speed increases from 00 to 10 m/s10\,m/s over the first 10 s10\,s. It remains constant at 10 m/s10\,m/s between 10 s10\,s and 15 s15\,s. Finally, the car decelerates from 10 m/s10\,m/s back to 00 by the time it reaches 20 s20\,s.

To work out the total distance covered by the car, one must calculate the total area under the graph. This can be viewed as a trapezoid. The area is calculated as: Distance=Area of Triangle 1+Area of Rectangle+Area of Triangle 2\text{Distance} = \text{Area of Triangle 1} + \text{Area of Rectangle} + \text{Area of Triangle 2}Distance=(12×10×10)+(5×10)+(12×5×10)\text{Distance} = (\frac{1}{2} \times 10 \times 10) + (5 \times 10) + (\frac{1}{2} \times 5 \times 10)Distance=50+50+25=125 m\text{Distance} = 50 + 50 + 25 = 125\,m

The average speed of the car for the entire journey is determined by dividing the total distance by the total time taken: Average Speed=125 m20 s=6.25 m/s\text{Average Speed} = \frac{125\,m}{20\,s} = 6.25\,m/s

The acceleration of the car during the initial phase is the slope of the first line: Acceleration=Change in SpeedTime Taken=10 m/s−0 m/s10 s=1.0 m/s2\text{Acceleration} = \frac{\text{Change in Speed}}{\text{Time Taken}} = \frac{10\,m/s - 0\,m/s}{10\,s} = 1.0\,m/s^2

Specific Speed-Time Calculation Scenarios

In another scenario, a car starts from rest and reaches a speed of 30 m/s30\,m/s after 15 s15\,s. The graph shows a straight diagonal line from the origin (0,0)(0, 0) to the point (15,30)(15, 30). At the start (0 s)(0\,s), the car's speed is 0 m/s0\,m/s. After 10 s10\,s, by observing the linear progression, the car's speed is 20 m/s20\,m/s. After the first 15 s15\,s, the graph identifies that the car reached a steady speed because the line becomes horizontal, indicating that the speed is no longer changing. This steady speed is exactly 30 m/s30\,m/s.

To find the distance travelled during the first 15 s15\,s, find the area of the triangle formed: Distance0−15=12×base×height=12×15 s×30 m/s=225 m\text{Distance}_{0-15} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 15\,s \times 30\,m/s = 225\,m

To find the total distance travelled during the first 30 s30\,s, add the area of the rectangle produced from 15 s15\,s to 30 s30\,s to the previous triangle: Area of Rectangle=(30 s−15 s)×30 m/s=15×30=450 m\text{Area of Rectangle} = (30\,s - 15\,s) \times 30\,m/s = 15 \times 30 = 450\,mTotal Distance=225 m+450 m=675 m\text{Total Distance} = 225\,m + 450\,m = 675\,m

Mechanics and Interpretation of Distance-Time Graphs

A distance-time graph illustrates the position of an object relative to a starting point over time. In these graphs, the vertical axis represents distance (usually in meters) and the horizontal axis represents time (usually in seconds). Slant lines (diagonal lines) indicate that the object is moving at a certain speed. A flat horizontal line indicates that the distance is not changing, meaning the car has stopped or is at rest.

Quantitative Analysis of a Non-Uniform Distance-Time Graph

A specific journey is recorded from point A to point D, where D is located 500 m500\,m from A. The journey is split into three phases: AB, BC, and CD.

Phase AB occurs during the first 10 s10\,s. The car moves from 0 m0\,m to 100 m100\,m. The speed during this phase is the gradient of the line: SpeedAB=DistanceTime=100 m10 s=10 m/s\text{Speed}_{AB} = \frac{\text{Distance}}{\text{Time}} = \frac{100\,m}{10\,s} = 10\,m/s

Phase BC represents the period where the car is stopped. The distance remains constant at 100 m100\,m from 10 s10\,s to 30 s30\,s. Therefore, the duration for which the car stopped is: Stop Duration=30 s−10 s=20 s\text{Stop Duration} = 30\,s - 10\,s = 20\,s

Phase CD shows the movement from point C (30 s,100 m)(30\,s, 100\,m) to point D (50 s,500 m)(50\,s, 500\,m). The speed for this segment is: SpeedCD=500 m−100 m50 s−30 s=400 m20 s=20 m/s\text{Speed}_{CD} = \frac{500\,m - 100\,m}{50\,s - 30\,s} = \frac{400\,m}{20\,s} = 20\,m/s

By comparing the speeds, the car travelled faster on the journey from C to D (20 m/s20\,m/s) compared to the journey from A to B (10 m/s10\,m/s).

Average Speed Calculations for Total and Moving Time

When evaluating the entire journey from point A to point D, two types of average speed can be calculated. The average speed for the whole journey considers all time elapsed, including the time spent at rest: Average Speedtotal=Total DistanceTotal Time=500 m50 s=10 m/s\text{Average Speed}_{total} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{500\,m}{50\,s} = 10\,m/s

Alternatively, the average speed for only the time the car was moving excludes the duration of the stop. The total time moving is the time from A to B (10 s10\,s) plus the time from C to D (20 s20\,s), resulting in a total of 30 s30\,s of motion: Average Speedmoving=Total DistanceTime Moving=500 m30 s=16.67 m/s\text{Average Speed}_{moving} = \frac{\text{Total Distance}}{\text{Time Moving}} = \frac{500\,m}{30\,s} = 16.67\,m/s