Velocity-Time Graphs

Physics

1. Fundamental Concepts

Definition: Velocity-Time Graphs represent the velocity of an object as a function of time.
Slope: The slope of the graph at any point gives the acceleration of the object.
Area Under Curve: The area under the curve between two points in time represents the displacement of the object during that interval.

2. Key Concepts

Constant Velocity: $v(t) = v_0$

Uniform Acceleration: $v(t) = v_0 + at$

Instantaneous Acceleration: $a = \frac{dv}{dt}$

3. Examples

Example 1 (Basic)

Problem: A car starts from rest and accelerates uniformly at $2 \text{{m/s}}^2$. Draw the velocity-time graph for the first $5$ seconds.

Step-by-Step Solution:

The initial velocity ($v_0$) is $0 \text{{m/s}}$.
The acceleration ($a$) is $2 \text{{m/s}}^2$.
The velocity at any time $t$ is given by $v(t) = 0 + 2t$.
Plot the line starting from $(0, 0)$ with a slope of $2$.

Validation: At $t = 5 \text{{s}}$, $v(5) = 2 \cdot 5 = 10 \text{{m/s}}$. ✓

Example 2 (Intermediate)

Problem: A particle has a velocity given by $v(t) = 4t - t^2$. Find the displacement from $t = 0$ to $t = 4$ seconds.

Step-by-Step Solution:

The velocity function is $v(t) = 4t - t^2$.
The displacement is the integral of velocity over time: $\int_{0}^{4} (4t - t^2) dt$.
Evaluate the integral: $\left[2t^2 - \frac{1}{3}t^3\right]_{0}^{4} = 2(4)^2 - \frac{1}{3}(4)^3 = 32 - \frac{64}{3} = \frac{96}{3} - \frac{64}{3} = \frac{32}{3}$.

Validation: The displacement from $t = 0$ to $t = 4$ is $\frac{32}{3} \text{{m}}$. ✓

4. Problem-Solving Techniques

Graphical Interpretation: Use graphs to visualize changes in velocity and acceleration.
Integration: For finding displacement, integrate the velocity function over the time interval.
Slope Analysis: Determine acceleration by calculating the slope of the velocity-time graph.