To solve this problem, we need to determine the expressions for the velocity $v(t)$ and the position $s(t)$ of an object given an initial velocity and an acceleration function.

Given Data:

1. Initial velocity, $v(0) = -30 \, \text{m/s}$
2. Acceleration function, $a(t) = 6t + 14 \, \text{m/s}^2$

Steps to Solve

1. Finding the Velocity $v(t)$:

Since acceleration $a(t)$ is the derivative of velocity with respect to time, we can integrate $a(t)$ to find $v(t)$.

$v(t) = \int a(t) \, dt = \int (6t + 14) \, dt$

Integrating $6t$ and $14$ separately: $v(t) = 6 \cdot \frac{t^2}{2} + 14t + C$ $v(t) = 3t^2 + 14t + C$

We use the initial condition $v(0) = -30$ to find $C$: $v(0) = 3(0)^2 + 14(0) + C = -30$ $C = -30$

Thus, the velocity function is: $v(t) = 3t^2 + 14t - 30$

2. Finding the Position $s(t)$:

Since velocity $v(t)$ is the derivative of position with respect to time, we integrate $v(t)$ to find $s(t)$.

$s(t) = \int v(t) \, dt = \int (3t^2 + 14t - 30) \, dt$

Integrating each term: $s(t) = 3 \cdot \frac{t^3}{3} + 14 \cdot \frac{t^2}{2} - 30t + C'$ $s(t) = t^3 + 7t^2 - 30t + C'$

Assuming the starting position is $s(0) = 0$, we use this to solve for $C'$: $s(0) = (0)^3 + 7(0)^2 - 30(0) + C' = 0$ $C' = 0$

Therefore, the position function is: $s(t) = t^3 + 7t^2 - 30t$

Final Answers

1. Velocity after $t$ seconds, $v(t)$: $v(t) = 3t^2 + 14t - 30 \, \text{m/s}$

2. Position after $t$ seconds, $s(t)$: $s(t) = t^3 + 7t^2 - 30t \, \text{m from the starting point}$

Would you like a deeper breakdown of any specific part of the solution, or do you have any further questions?

Related Questions to Explore

1. What would happen to the velocity function if the initial velocity were different?
2. How would the position equation change if acceleration was a constant rather than a function of $t$?
3. How can we interpret the physical significance of the constants in the velocity and position equations?
4. What is the object's velocity and position at $t = 5$ seconds?
5. How can we determine when the object changes direction?

Tip:

In problems involving acceleration, velocity, and position, remember that integrating gives you the function for the next higher derivative (e.g., integrating acceleration yields velocity, and integrating velocity yields position).