To find a function describing the position of the particle, we'll need to integrate the given acceleration function twice.
Let's start by integrating the acceleration function with respect to time:
∫a(t) dt = ∫(10sin(t) + 3cos(t)) dt
To integrate sin(t), we use the identity ∫sin(t) dt = -cos(t) + C.
To integrate cos(t), we use the identity ∫cos(t) dt = sin(t) + C.
Using these identities, we can rewrite the integral as:
∫(10sin(t) + 3cos(t)) dt = -10cos(t) + 3sin(t) + C₁
Now, we have the velocity function v(t), which is the integral of the acceleration function:
v(t) = -10cos(t) + 3sin(t) + C₁
Next, we'll need to find the constant C₁. To do that, we can use the initial condition s(0) = -4. Since s(t) is the integral of v(t), we can find s(t) by integrating v(t) and applying the initial condition.
Integrating v(t) with respect to time, we get:
∫v(t) dt = -∫(10cos(t) - 3sin(t)) dt
Using the same trigonometric identities as before, we can rewrite the integral as:
∫v(t) dt = -10sin(t) - 3cos(t) + C₂
Now, we have the position function s(t), which is the integral of v(t):
s(t) = -10sin(t) - 3cos(t) + C₂
To find the constant C₂, we can use the other initial condition s(2pi) = 1. Plugging in t = 2π into the position function, we can solve for C₂:
-10sin(2π) - 3cos(2π) + C₂ = 1
0 - 3 + C₂ = 1
C₂ = 4
Now we have the final position function:
s(t) = -10sin(t) - 3cos(t) + 4
Therefore, the function describing the position of the particle is s(t) = -10sin(t) - 3cos(t) + 4.