Question:
A force F = 6t acts on a body of mass 2 kg initially at rest. Find velocity at t = 2 s.
A force F = 6t acts on a body of mass 2 kg initially at rest. Find velocity at t = 2 s.
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Never use the standard equations of motion (like $v = u + at$) when the force is variable. You must always use calculus (integration) because acceleration changes with time.
Updated On: Jun 3, 2026
- 3 m/s
- 6 m/s
- 9 m/s
- 12 m/s
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Question:
We are given a time-varying force $F = 6t$ acting on a body of mass $2 \text{ kg}$ which starts from rest (initial velocity $u = 0$ at $t = 0$). We need to determine the final velocity of the body at $t = 2 \text{ s}$.
Step 2: Key Formula or Approach:
According to Newton's Second Law of Motion, force is mass times acceleration: \[ F = m a = m \frac{dv}{dt} \] We can rearrange this differential equation to solve for the velocity: \[ dv = \frac{F}{m} \, dt \] Integrating both sides from $t = 0$ to $t = t$: \[ \int_{0}^{v} dv = \int_{0}^{t} \frac{F}{m} \, dt \]
Step 3: Detailed Explanation:
• Identify the given terms:
Mass ($m$) = $2 \text{ kg}$
Force ($F$) = $6t$
Initial state: $v = 0$ at $t = 0$
• Calculate the acceleration ($a$) as a function of time:
\[ a(t) = \frac{F}{m} = \frac{6t}{2} = 3t \text{ m/s}^2 \]
• Set up the integration for the velocity at $t = 2 \text{ s}$:
\[ v = \int_{0}^{2} a(t) \, dt = \int_{0}^{2} 3t \, dt \]
• Perform the integration:
\[ v = \left[ \frac{3t^2}{2} \right]_{0}^{2} \]
\[ v = \frac{3 \times 2^2}{2} - 0 = \frac{3 \times 4}{2} = 6 \text{ m/s} \]
• The final calculated velocity of the body at $t = 2 \text{ s}$ is exactly $6 \text{ m/s}$.
Step 4: Final Answer:
The velocity of the body at $t = 2 \text{ s}$ is $6 \text{ m/s}$.
We are given a time-varying force $F = 6t$ acting on a body of mass $2 \text{ kg}$ which starts from rest (initial velocity $u = 0$ at $t = 0$). We need to determine the final velocity of the body at $t = 2 \text{ s}$.
Step 2: Key Formula or Approach:
According to Newton's Second Law of Motion, force is mass times acceleration: \[ F = m a = m \frac{dv}{dt} \] We can rearrange this differential equation to solve for the velocity: \[ dv = \frac{F}{m} \, dt \] Integrating both sides from $t = 0$ to $t = t$: \[ \int_{0}^{v} dv = \int_{0}^{t} \frac{F}{m} \, dt \]
Step 3: Detailed Explanation:
• Identify the given terms:
Mass ($m$) = $2 \text{ kg}$
Force ($F$) = $6t$
Initial state: $v = 0$ at $t = 0$
• Calculate the acceleration ($a$) as a function of time:
\[ a(t) = \frac{F}{m} = \frac{6t}{2} = 3t \text{ m/s}^2 \]
• Set up the integration for the velocity at $t = 2 \text{ s}$:
\[ v = \int_{0}^{2} a(t) \, dt = \int_{0}^{2} 3t \, dt \]
• Perform the integration:
\[ v = \left[ \frac{3t^2}{2} \right]_{0}^{2} \]
\[ v = \frac{3 \times 2^2}{2} - 0 = \frac{3 \times 4}{2} = 6 \text{ m/s} \]
• The final calculated velocity of the body at $t = 2 \text{ s}$ is exactly $6 \text{ m/s}$.
Step 4: Final Answer:
The velocity of the body at $t = 2 \text{ s}$ is $6 \text{ m/s}$.
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