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.