Question:
The velocity-time graph of a particle is given by v = 4t. The acceleration is:
The velocity-time graph of a particle is given by v = 4t. The acceleration is:
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For any linear velocity equation of the form $v = k t$, the acceleration is simply the constant coefficient $k$. Here, since $v = 4t$, the acceleration is immediately $4 \text{ m/s}^2$.
Updated On: May 21, 2026
- 2 m/$s^2$
- 4 m/$s^2$
- 8 m/$s^2$
- Variable
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Question:
The question provides a linear relationship for velocity ($v$) as a function of time ($t$) given by $v = 4t$. We are required to find the acceleration of the particle from this relationship.
Step 2: Key Formula or Approach:
Acceleration ($a$) is defined as the time rate of change of velocity: \[ a = \frac{dv}{dt} \] In a velocity-time graph, the acceleration corresponds to the slope of the curve.
Step 3: Detailed Explanation:
• Given the velocity equation:
\[ v(t) = 4t \]
• To find the acceleration, we differentiate the velocity function with respect to time ($t$):
\[ a = \frac{d}{dt}(4t) \]
• Applying the power rule of differentiation ($\frac{d}{dt}(t) = 1$):
\[ a = 4 \times 1 = 4 \text{ m/s}^2 \]
• Because the derivative is a constant numerical value ($4$), it does not depend on time ($t$). This tells us that the acceleration is uniform and constant.
• Thus, the particle moves with a constant acceleration of $4 \text{ m/s}^2$.
Step 4: Final Answer:
The acceleration of the particle is constant and has a magnitude of $4 \text{ m/s}^2$.
The question provides a linear relationship for velocity ($v$) as a function of time ($t$) given by $v = 4t$. We are required to find the acceleration of the particle from this relationship.
Step 2: Key Formula or Approach:
Acceleration ($a$) is defined as the time rate of change of velocity: \[ a = \frac{dv}{dt} \] In a velocity-time graph, the acceleration corresponds to the slope of the curve.
Step 3: Detailed Explanation:
• Given the velocity equation:
\[ v(t) = 4t \]
• To find the acceleration, we differentiate the velocity function with respect to time ($t$):
\[ a = \frac{d}{dt}(4t) \]
• Applying the power rule of differentiation ($\frac{d}{dt}(t) = 1$):
\[ a = 4 \times 1 = 4 \text{ m/s}^2 \]
• Because the derivative is a constant numerical value ($4$), it does not depend on time ($t$). This tells us that the acceleration is uniform and constant.
• Thus, the particle moves with a constant acceleration of $4 \text{ m/s}^2$.
Step 4: Final Answer:
The acceleration of the particle is constant and has a magnitude of $4 \text{ m/s}^2$.
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