To find the vector r(t) (position vector as a function of time), we need to relate it to the acceleration vector. Given that acceleration a(t) is constant at $3 \ \text{m/s}^2$, we can integrate to find the velocity vector v(t) and then integrate again to find the position vector r(t).

Step-by-step solution:

1. Acceleration $a(t)$: $a(t) = 3 \ \text{m/s}^2$ This implies that acceleration is constant.

2. Velocity $v(t)$: To find velocity, integrate acceleration with respect to time: $v(t) = \int a(t) \, dt = \int 3 \, dt = 3t + C_1$ where $C_1$ is the constant of integration, representing the initial velocity $v_0$.

3. Position $r(t)$: Now, integrate the velocity function with respect to time: $r(t) = \int v(t) \, dt = \int (3t + C_1) \, dt = \frac{3t^2}{2} + C_1 t + C_2$ where $C_2$ is another constant of integration, representing the initial position $r_0$.

Final expression for $r(t)$:

$r(t) = \frac{3t^2}{2} + v_0 t + r_0$ Here:

• $v_0$ is the initial velocity at $t = 0$,
• $r_0$ is the initial position at $t = 0$.

To fully determine $r(t)$, you would need the initial conditions: $v_0$ and $r_0$.

Would you like more details on this? Or perhaps clarification on the integration process?

Related Questions:

1. How do you find the velocity when the acceleration is non-constant?
2. What are the steps for solving kinematic problems with initial velocity and acceleration?
3. How do you interpret the constants $v_0$ and $r_0$ in real-world problems?
4. How would the result change if acceleration was a vector instead of a scalar?
5. How do you graph position vs. time when acceleration is constant?

Tip:

When solving kinematic problems, always check for initial conditions like starting velocity and position to fully determine the motion.