Let's solve the problem step-by-step.

Given Information:

1. $f(0) = 6$ (initial displacement).
2. $f''(t) = 12t$ (acceleration).
3. $f'(0) = 1$ (initial velocity).

Step 1: Find $f'(t)$ (velocity)

The acceleration $f''(t) = 12t$ is the derivative of velocity. Integrate $f''(t)$ with respect to $t$:

$f'(t) = \int f''(t) \, dt = \int 12t \, dt = 6t^2 + C_1,$ where $C_1$ is the constant of integration.

Using $f'(0) = 1$: $f'(0) = 6(0)^2 + C_1 = 1 \implies C_1 = 1.$

Thus, the velocity function is: $f'(t) = 6t^2 + 1.$

Step 2: Find $f(t)$ (displacement)

The velocity $f'(t) = 6t^2 + 1$ is the derivative of displacement. Integrate $f'(t)$ with respect to $t$:

$f(t) = \int f'(t) \, dt = \int (6t^2 + 1) \, dt = 2t^3 + t + C_2,$ where $C_2$ is the constant of integration.

Using $f(0) = 6$: $f(0) = 2(0)^3 + 0 + C_2 = 6 \implies C_2 = 6.$

Thus, the displacement function is: $f(t) = 2t^3 + t + 6.$

Step 3: Find $f(1)$ (displacement at $t = 1$)

Substitute $t = 1$ into $f(t)$: $f(1) = 2(1)^3 + (1) + 6 = 2 + 1 + 6 = 9.$

Final Answer:

1. The equation of displacement is: $f(t) = 2t^3 + t + 6.$
2. The displacement of the particle at $t = 1$ is: $f(1) = 9.$

Would you like me to explain a step in more detail or verify further?

Related Questions:

1. How is acceleration related to displacement through integration?
2. What if the initial velocity ($f'(0)$) was zero? How would it affect the result?
3. Can we find the time when the particle’s velocity becomes zero?
4. How do we confirm that the integration constants were determined correctly?
5. What does the graph of $f(t)$ tell us about the motion?

Tip:

When solving problems involving integration, always carefully include constants of integration and use initial conditions to find their values.