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The problem provides the acceleration function \( a(t) = \frac{2t}{(t^2 + 1)^2} \), along with the initial conditions \( v(0) = 0 \) (initial velocity) and \( s(0) = 0 \) (initial position).

To find the velocity \( v(t) \) and the position \( s(t) \), we need to integrate the acceleration.

### Step 1: Find Velocity \( v(t) \)
The velocity is the integral of the acceleration:
\[
v(t) = \int a(t) \, dt = \int \frac{2t}{(t^2 + 1)^2} \, dt
\]

This is a standard integral. We can solve it using substitution:
- Let \( u = t^2 + 1 \), then \( du = 2t \, dt \).
Thus, the integral becomes:
\[
v(t) = \int \frac{1}{u^2} \, du = -\frac{1}{u} + C = -\frac{1}{t^2 + 1} + C
\]

Now, using the initial condition \( v(0) = 0 \), we substitute \( t = 0 \):
\[
0 = -\frac{1}{0^2 + 1} + C \quad \Rightarrow \quad C = 1
\]
Thus, the velocity function is:
\[
v(t) = 1 - \frac{1}{t^2 + 1}
\]

### Step 2: Find Position \( s(t) \)
The position is the integral of the velocity:
\[
s(t) = \int v(t) \, dt = \int \left( 1 - \frac{1}{t^2 + 1} \right) \, dt
\]
This separates into two integrals:
\[
s(t) = \int 1 \, dt - \int \frac{1}{t^2 + 1} \, dt
\]
The first integral is \( t \), and the second is \( \arctan(t) \):
\[
s(t) = t - \arctan(t) + C
\]

Using the initial condition \( s(0) = 0 \), we substitute \( t = 0 \):
\[
0 = 0 - \arctan(0) + C \quad \Rightarrow \quad C = 0
\]
Thus, the position function is:
\[
s(t) = t - \arctan(t)
\]

### Final Answer:
- Velocity: \( v(t) = 1 - \frac{1}{t^2 + 1} \)
- Position: \( s(t) = t - \arctan(t) \)

Would you like further details or have any questions?

Here are 5 related questions to expand your understanding:
1. What is the significance of the initial conditions \( v(0) = 0 \) and \( s(0) = 0 \)?
2. How can substitution be applied to more complicated integrals?
3. What is the geometric interpretation of the function \( \arctan(t) \)?
4. How would the solution change if \( v(0) \) was a non-zero value?
5. What happens to the velocity and position as \( t \to \infty \)?

**Tip:** When integrating, always carefully check the boundaries and initial conditions to find the constant of integration.

The problem provides the acceleration function a(t) = 2t / (t^2 + 1)^2, along with the initial conditions v(0) = 0 (initial velocity) and s(0) = 0 (initial position).

This problem demonstrates how to find velocity and position functions by integrating the acceleration function a(t) = 2t / (t^2 + 1)^2, using initial conditions v(0) = 0 and s(0) = 0. The solution involves substitution in integration and the application of calculus principles.

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