Let's break down the problem step by step.

1. Given information:

• The particle has mass $m$.
• It starts at rest, so $v(0) = 0$ and $x(0) = 0$.
• The force acting on the particle increases with time as $F(t) = Ct^2$, where $C$ is a constant.
• We need to determine the velocity $v(t)$ and position $x(t)$ as functions of time.

2. Newton's Second Law:

Newton's Second Law relates force to acceleration: $F(t) = m a(t)$ where $a(t)$ is the acceleration of the particle. We also know that $a(t) = \frac{dv(t)}{dt}$.

Given that $F(t) = Ct^2$, we can substitute this into Newton's Second Law: $m a(t) = Ct^2$ $m \frac{dv(t)}{dt} = Ct^2$ Now, solve for $\frac{dv(t)}{dt}$: $\frac{dv(t)}{dt} = \frac{C}{m} t^2$

3. Integrate to find $v(t)$:

Integrating both sides with respect to time $t$: $v(t) = \int \frac{C}{m} t^2 \, dt$ $v(t) = \frac{C}{m} \int t^2 \, dt$ $v(t) = \frac{C}{m} \left( \frac{t^3}{3} \right) + v(0)$ Since the particle starts from rest, $v(0) = 0$. So: $v(t) = \frac{C}{3m} t^3$

4. Determine the position $x(t)$:

We know that velocity is the time derivative of position: $v(t) = \frac{dx(t)}{dt}$. Therefore: $\frac{dx(t)}{dt} = \frac{C}{3m} t^3$ Now, integrate this to find $x(t)$: $x(t) = \int \frac{C}{3m} t^3 \, dt$ $x(t) = \frac{C}{3m} \int t^3 \, dt$ $x(t) = \frac{C}{3m} \left( \frac{t^4}{4} \right) + x(0)$ Since $x(0) = 0$, the final expression for $x(t)$ is: $x(t) = \frac{C}{12m} t^4$

Summary:

• The velocity of the particle as a function of time is: $v(t) = \frac{C}{3m} t^3$

• The position of the particle as a function of time is: $x(t) = \frac{C}{12m} t^4$

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Would you like more details or have any questions?

Here are 5 related questions for further exploration:

1. How would the result change if the force was $F = C t$ instead of $F = C t^2$?
2. What is the acceleration $a(t)$ as a function of time for this force?
3. How would you find the time when the particle reaches a specific velocity?
4. How would you calculate the work done by the force up to time $t$?
5. What if the particle had an initial velocity? How would that change the equations?

Tip: When dealing with time-dependent forces, always use Newton's Second Law and integrate step by step to find velocity and position.