𝑎(𝑡)=8.7𝑡2+6,where 𝑡 is measured in seconds.1. Write an equation for the velocity 𝑣 at time 𝑡. Simplify your answer.𝑣(𝑡)=

To find the velocity $v(t)$ at time $t$ given the acceleration $a(t) = 8.7t^2 + 6$, we need to integrate the acceleration function with respect to time. Here are the steps:

1. Write down the given acceleration function: $a(t) = 8.7t^2 + 6$

2. Recall that velocity is the integral of acceleration: $v(t) = \int a(t) \, dt$

3. Set up the integral: $v(t) = \int (8.7t^2 + 6) \, dt$

4. Integrate each term separately:

   • The integral of $8.7t^2$ with respect to $t$ is: $\int 8.7t^2 \, dt = 8.7 \int t^2 \, dt = 8.7 \left( \frac{t^3}{3} \right) = 2.9t^3$
   • The integral of $6$ with respect to $t$ is: $\int 6 \, dt = 6t$

5. Combine the results and add the constant of integration $C$: $v(t) = 2.9t^3 + 6t + C$

So, the equation for the velocity $v(t)$ at time $t$ is: $v(t) = 2.9t^3 + 6t + C$

Here, $C$ is the constant of integration, which can be determined if an initial condition (such as the initial velocity) is provided.