Graphs (horizontal and vertical lines at equal distances) are an efficient method to show the relationship between physical quantities. Graphs use coordinate systems to illustrate the relationship between various quantities.
The distance-time graph is plotted between distance (s) and time (t). Time is plotted along the horizontal axis (x-axis), and distance is plotted along the vertical axis (y-axis) as shown in figure 2.12 (a). The distance-time graph is always in the positive X plane, as with the passage of time, distance never goes to negative values, irrespective of the direction of motion.
The gradient (or slope) of the distance-time curve gives speed. The gradient of the graph is calculated as:
From equation 2.1, it is defined that the gradient = v (speed). Thus, by looking at the graph, we get an idea about the speed of a body, shown in figure 2.13.
A car travels from Peshawar to Islamabad on Motorway (M1), stopping for 30 minutes at a 'rest area'. Calculate the speed of the car in km/hr and m/s for:
From the graph
(a) Distance between Peshawar to rest area is Δs = 100 km
Time taken from Peshawar to rest area is Δt = 1 hr
Gradient = v = Δs / Δt = 100 km / 1 hr = 100 km/hr
In meter per second: Δs = 100 km = 100,000 m and Δt = 1 hr = 3600 s
Gradient = v = 100,000 m / 3600 s = 27.78 m/s
Answer: 27.78 m/s
(b) Distance between rest area and Islamabad is Δs = 75 km
Time taken is Δt = 1 hr 30 mins = 1.5 hr
Gradient = v = Δs / Δt = 75 km / 1.5 hr = 50 km/hr
In meter per second: Δs = 75 km = 75,000 m and Δt = 1.5 hr = 5400 s
Gradient = v = 75,000 m / 5400 s = 13.89 m/s
Answer: 13.89 m/s
(c) Distance from Peshawar to Islamabad is Δs = 175 km
Time taken is Δt = 2.5 hr
Gradient = v = Δs / Δt = 175 km / 2.5 hr = 70 km/hr
In meter per second: Δs = 175 km = 175,000 m and Δt = 2.5 hr = 9000 s
Gradient = v = 175,000 m / 9000 s = 19.44 m/s
Answer: 19.44 m/s
The speed-time graph is plotted between speed (v) and time (t). In this graphical analysis, speed is plotted along the vertical axis (y-axis), and time along the horizontal axis (x-axis). Speed-time graphs serve two purposes:
The slope of the speed-time curve gives the magnitude of acceleration. The gradient (or slope) of the speed-time curve is calculated as:
The area under the speed-time graph represents the distance traveled. If the motion of a body is represented by a geometric shape, the area can be calculated using appropriate formulas.
For example, if the motion is represented by a rectangle, the area of the rectangle is given by:
Area = width × length = v × t
Similarly, if the graph is represented by a triangle, the area of the triangle is given by:
Area = 0.5 × width × length = 0.5 × v × t
A car increases its speed from zero to 30 m/s in 20 s. Then it moves with uniform speed for the next 30 seconds, and then the driver applies brakes, causing the speed to decrease uniformly to zero in 10 s. Use this graph to calculate:
(a) The slope of the graph will give the magnitude of acceleration.
(i) For the first 20 seconds, the slope OA is:
Magnitude of acceleration = Δv / Δt = (30 - 0) m/s / (20 - 0) s = 1.5 m/s²
Answer: 1.5 m/s²
(ii) From 20 s to 50 s, the slope AB is:
Magnitude of acceleration = Δv / Δt = (30 - 30) m/s / (50 - 20) s = 0 m/s²
Answer: 0 m/s²
(iii) In the last 10 seconds, the slope BC is:
Magnitude of acceleration = Δv / Δt = (0 - 30) m/s / (60 - 50) s = -3 m/s²
Answer: -3 m/s² (negative sign indicates slowing down)
(b) Total distance covered is equal to the area under the speed-time graph:
Total distance = Area of triangle OAE + Area of rectangle ABDE + Area of triangle CBD
= (0.5 × 30 m/s × 20 s) + (30 m/s × 30 s) + (0.5 × 30 m/s × 10 s)
= 300 m + 900 m + 150 m = 1350 m
Answer: 1350 m
(c) Average speed = Total distance / Total time
= 1350 m / 60 s = 22.5 m/s
Answer: 22.5 m/s
What does the slope of these graphs tell about acceleration?