Give a detailed explanation on how to solve:  a ball moves in a straight line an has an accelaration of a(t)=2t+5. Find the position function of the ball if its initial velocity is -3 cm/sec and its initial position is 12 cm.

To solve the problem of finding the position function \( s(t) \) of a ball given its acceleration function \( a(t) = 2t + 5 \), initial velocity, and initial position, we follow a step-by-step approach involving integration. 

### Step 1: Understand the Given Information
- **Acceleration function:** \( a(t) = 2t + 5 \)
- **Initial velocity:** \( v(0) = -3 \) cm/sec
- **Initial position:** \( s(0) = 12 \) cm

### Step 2: Find the Velocity Function
The velocity function \( v(t) \) is the integral of the acceleration function \( a(t) \).

\[
v(t) = \int a(t) \, dt = \int (2t + 5) \, dt
\]

To integrate \( 2t + 5 \), we apply the power rule for integration:

\[
\int 2t \, dt = t^2 \quad \text{and} \quad \int 5 \, dt = 5t
\]

Thus,

\[
v(t) = t^2 + 5t + C_1
\]

where \( C_1 \) is the constant of integration. 

### Step 3: Determine the Constant \( C_1 \)
To find \( C_1 \), use the initial velocity condition \( v(0) = -3 \):

\[
v(0) = (0)^2 + 5(0) + C_1 = C_1
\]

Given that \( v(0) = -3 \):

\[
C_1 = -3
\]

So the velocity function is:

\[
v(t) = t^2 + 5t - 3
\]

### Step 4: Find the Position Function
The position function \( s(t) \) is the integral of the velocity function \( v(t) \):

\[
s(t) = \int v(t) \, dt = \int (t^2 + 5t - 3) \, dt
\]

Integrating term by term:

\[
\int t^2 \, dt = \frac{t^2 \cdot t}{3} = \frac{t^3}{3}, \quad \int 5t \, dt = \frac{5t^2}{2}, \quad \text{and} \quad \int -3 \, dt = -3t
\]

Thus,

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

where \( C_2 \) is another constant of integration.

### Step 5: Determine the Constant \( C_2 \)
To find \( C_2 \), use the initial position condition \( s(0) = 12 \):

\[
s(0) = \frac{(0)^3}{3} + \frac{5(0)^2}{2} - 3(0) + C_2 = C_2
\]

Given that \( s(0) = 12 \):

\[
C_2 = 12
\]

So the position function is:

\[
s(t) = \frac{t^3}{3} + \frac{5t^2}{2} - 3t + 12
\]

### Final Answer
The position function of the ball is:

\[
\boxed{s(t) = \frac{t^3}{3} + \frac{5t^2}{2} - 3t + 12}
\]

This function gives the position of the ball at any time \( t \).

---

Would you like any further details or have any questions? Here are some related questions you might find helpful:

1. How would the position function change if the initial velocity were different?
2. What is the velocity of the ball at \( t = 2 \) seconds?
3. Can you find the time at which the ball comes to rest?
4. How do you find the maximum height (position) reached by the ball?
5. What does the acceleration function tell us about the motion of the ball over time?

**Tip:** When dealing with motion problems, remember that the relationship between acceleration, velocity, and position is rooted in integration and differentiation.

Discover how to find the position function of a ball moving with acceleration 2t + 5, given initial velocity and position. This detailed solution covers integration of acceleration to find velocity and further integration to determine position. Suitable for high school students studying calculus and kinematics.

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