Which graph represents positive acceleration ?

Options

A.
B.
C.
D.

Answer

To determine which graph represents positive acceleration, we need to analyze the velocity-time graphs provided in the question. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Concept of Acceleration
Acceleration is defined as the rate of change of velocity with respect to time. Mathematically, it is given by:
\[ A = \frac{dV}{dt} \]
where \( A \) is acceleration, \( V \) is velocity, and \( t \) is time.

### Step 2: Analyze Each Graph
We will analyze each graph one by one to determine the acceleration.

#### Graph 1:
- **Observation**: The velocity (V) is constant at 0.
- **Calculation**: Since the velocity is constant, the change in velocity (\( dV \)) is 0.
- **Conclusion**:
\[ A = \frac{dV}{dt} = \frac{0}{dt} = 0 \]
Therefore, this graph represents zero acceleration.

#### Graph 2:
- **Observation**: The velocity (V) is constant (not changing).
- **Calculation**: Similar to the first graph, since the velocity is constant, \( dV \) is again 0.
- **Conclusion**:
\[ A = \frac{dV}{dt} = 0 \]
This graph also represents zero acceleration.

#### Graph 3:
- **Observation**: The velocity starts at 0 and decreases to a negative value.
- **Calculation**: Let’s denote the initial velocity \( V_1 = 0 \) at time \( T_1 = 0 \) and final velocity \( V_2 = -V \) at time \( T_2 = T \).
\[ A = \frac{V_2 - V_1}{T_2 - T_1} = \frac{-V - 0}{T - 0} = \frac{-V}{T} \]
- **Conclusion**: Since \( A \) is negative, this graph represents negative acceleration.

#### Graph 4:
- **Observation**: The velocity starts at a negative value and increases to 0.
- **Calculation**: Let’s denote the initial velocity \( V_1 = -V \) at time \( T_1 = 0 \) and final velocity \( V_2 = 0 \) at time \( T_2 = T \).
\[ A = \frac{V_2 - V_1}{T_2 - T_1} = \frac{0 - (-V)}{T - 0} = \frac{V}{T} \]
- **Conclusion**: Since \( A \) is positive, this graph represents positive acceleration.

### Final Conclusion:
The graph that represents positive acceleration is **Graph 4**.

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