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The problem you're tackling involves a particle moving along a straight line, and its acceleration is given as a function of time \( t \):

\[
a(t) = 12t - 3t^{1/2} \quad \text{in} \quad \text{feet}/\text{second}^2.
\]

We are asked to determine both the velocity and position of the particle as functions of time, with the initial conditions:

- \( t = 0 \)
- \( v = 0 \) (velocity at \( t = 0 \))
- \( s = 15 \) feet (position at \( t = 0 \)).

### Step 1: Find the velocity function

Since acceleration \( a(t) \) is the derivative of velocity \( v(t) \), we integrate the acceleration function to find the velocity:

\[
v(t) = \int a(t) \, dt = \int (12t - 3t^{1/2}) \, dt
\]

Performing the integration:

\[
v(t) = \int 12t \, dt - \int 3t^{1/2} \, dt
\]
\[
v(t) = 6t^2 - 2t^{3/2} + C_1
\]

Here, \( C_1 \) is the constant of integration, which we will determine using the initial condition \( v(0) = 0 \):

\[
v(0) = 6(0)^2 - 2(0)^{3/2} + C_1 = 0 \quad \Rightarrow \quad C_1 = 0
\]

Thus, the velocity function is:

\[
v(t) = 6t^2 - 2t^{3/2}
\]

### Step 2: Find the position function

Next, to find the position function \( s(t) \), we integrate the velocity function:

\[
s(t) = \int v(t) \, dt = \int (6t^2 - 2t^{3/2}) \, dt
\]

Performing the integration:

\[
s(t) = 2t^3 - \frac{4}{5}t^{5/2} + C_2
\]

Here, \( C_2 \) is another constant of integration, which we determine using the initial condition \( s(0) = 15 \):

\[
s(0) = 2(0)^3 - \frac{4}{5}(0)^{5/2} + C_2 = 15 \quad \Rightarrow \quad C_2 = 15
\]

Thus, the position function is:

\[
s(t) = 2t^3 - \frac{4}{5}t^{5/2} + 15
\]

### Final Answers

- The velocity as a function of time is:

\[
v(t) = 6t^2 - 2t^{3/2}
\]

- The position as a function of time is:

\[
s(t) = 2t^3 - \frac{4}{5}t^{5/2} + 15
\]

Would you like more details on any part of the solution or clarification on how the integration was done?

Here are 5 related questions you can explore:
1. What happens to the particle's velocity as \( t \to \infty \)?
2. How would the problem change if the acceleration function were different?
3. How do we interpret the negative term in the velocity function?
4. How can we verify the correctness of these solutions using differentiation?
5. How does the particle's position change after \( t = 5 \) seconds?

**Tip:** When integrating functions with fractional exponents, remember that the integral of \( t^n \) is \( \frac{t^{n+1}}{n+1} \), provided \( n \neq -1 \).

Una partícula se desplaza a lo largo de una línea recta con la aceleración a = (12t - 3t^(1/2)) pies/s^2, donde t está en segundos. Determine su velocidad y posición como una función del tiempo. Cuando t = 0, v = 0 y s = 15 pies.

This problem involves finding the velocity and position of a particle given an acceleration function a(t) = 12t - 3t^(1/2) in feet/s^2. The problem applies integral calculus, with initial conditions v(0) = 0 and s(0) = 15 feet, providing a step-by-step solution for physics and calculus students.

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