The variation in velocity (v) with time (t) for an object moving in a straight line can be represented by v-t graph. Area under v-t graph represents a physical quantity which has the unit-
Velocity-Time Graph Overview
A v-t graph, also known as a velocity-time graph, is a powerful tool in kinematics that visually illustrates how the velocity (v) of an object changes over a specific period of time (t). For an object moving in a straight line, this graph provides comprehensive information about its motion. The horizontal axis represents time, typically in seconds, while the vertical axis represents velocity, typically in metres per second.
Area Under v-t Graph Representation
The area under the v-t graph has significant physical meaning for an object's motion. When we calculate this area, we are essentially multiplying the velocity by the time. The physical quantity represented by the area under a velocity-time graph for an object moving in a straight line is its displacement. Displacement is a vector quantity that describes the change in position of an object, indicating both how far it has moved and in what direction.
The fundamental relationship that links velocity, time, and displacement is given by:
$$ \text{Displacement} = \text{Velocity} \times \text{Time} $$
This relationship is precisely what the area under the curve on a v-t graph represents. For example, if the velocity is constant, the area forms a rectangle, where one side is the velocity value and the other is the time interval. The product of these two dimensions gives the displacement.
Unit of Area Under v-t Graph
To determine the unit of the physical quantity represented by the area under the v-t graph, we multiply the units of the quantities plotted on the axes:
- The standard unit for velocity (v) is metre/second (\(\text{m/s}\)).
- The standard unit for time (t) is second (\(\text{s}\)).
Therefore, when we calculate the area, we are effectively performing a unit multiplication:
$$ \text{Unit of Area} = \text{Unit of Velocity} \times \text{Unit of Time} $$ $$ \text{Unit of Area} = \left( \frac{\text{metre}}{\text{second}} \right) \times \text{second} $$ $$ \text{Unit of Area} = \text{metre} $$
This calculation clearly shows that the physical quantity represented by the area under the v-t graph has the unit of metre, which correctly corresponds to displacement.
Analyzing the Options for Unit
Let's carefully consider each option provided in the context of the area under a v-t graph and its resulting unit:
| Option | Unit | Associated Physical Quantity | Relevance to Area Under v-t Graph |
|---|---|---|---|
| 1 | metre | Displacement, Distance, Length | This is the correct unit for the physical quantity (displacement) represented by the area under a velocity-time graph, as derived from \( \text{velocity} \times \text{time} \). |
| 2 | metre2 | Area (e.g., surface area of a shape) | This unit represents area in a two-dimensional geometric sense (like square metres for land), not the physical quantity (displacement) obtained from a velocity-time graph. |
| 3 | metre/second | Velocity, Speed | This is the unit of velocity (v) itself, which is plotted on the vertical axis of the v-t graph. It does not represent the area beneath the graph. |
| 4 | metre/second2 | Acceleration | This is the unit for acceleration. On a v-t graph, acceleration is represented by the slope of the graph, not the area under it. |
Conclusion on Area Unit
Based on the analysis of the units and the principles of kinematics, the physical quantity represented by the area under the v-t graph for an object moving in a straight line is displacement, and its corresponding unit is the metre.
Important Questions from Motion in a Straight Line
- Which of the following examples represents motion along a straight line?
Ram records the odometer readings of his car for the distance covered from 2000 km at the start of his journey and 2400 km at the end of the journey after 8 hours. What is the average speed of the car?
A body moves along a straight line with displacement \(s = 4t^2 + 2ts =4t^2+2ts=4t^2+2t\) (in m, t in s). Its instantaneous acceleration is:
The area under a velocity-time graph between two instants represents:
A body is dropped freely from a height 'X' m. If tit takes Y seconds to reach the ground. The time it takes to reach half of the height of the tower is (Take g = 10m/s2)
