Let s(t) denote the position of a particle at time t, and let v and a be the velocity and acceleration respectively. The particle is moving according to the data

a(t)=10sin(t)+3cos(t)
s(0)=-4
s(2pi)=1

find a function describing position of particle

s(t)=???

I do not know where to start, it is unlike other anti-derivative problems I have dealt with. I would prefer an explanation with an answer, thanks.

a(t) is the 2nd derivative of s(t)

v(t) is the 1st derivative of s(t)

take the anti-derivative of a(t)
... this is v(t) + C
... don't forget the unknown constant

take the anti-derivative of v(t)
... this is s(t) + Ct + D
... don't forget the unknown constants

use the two given values for s(t) to find the constants

Well, well, well, looks like we have a particle that's going on a joyride with some interesting acceleration! Don't worry, I'll help you navigate through this problem with a bit of humor.

First things first, let's integrate the given acceleration function to find the velocity function. The integral of sin(t) is -cos(t), and the integral of cos(t) is sin(t), right? Remember those trig identities!

So, we have:

v(t) = ∫(10sin(t) + 3cos(t)) dt
v(t) = -10cos(t) + 3sin(t) + C

Where C is the constant of integration. Now, we need to find the value of C using the given initial position.

Given: s(0) = -4

To find C, we plug in t = 0 into the velocity equation:

v(0) = -10cos(0) + 3sin(0) + C
v(0) = -10(1) + 3(0) + C
C = -10

Alright, now that we have the value of C, we can find the position function by integrating the velocity function. Get ready for the funny part!

s(t) = ∫(-10cos(t) + 3sin(t) - 10) dt
s(t) = -10sin(t) - 3cos(t) - 10t + D

Here, D is another constant of integration. To find D, we use the other piece of given information:

Given: s(2π) = 1

Plug in t = 2π into the position equation:

s(2π) = -10sin(2π) - 3cos(2π) - 10(2π) + D
1 = 0 - 3(1) - 20π + D
1 = -3 - 20π + D

Now, let's solve for D:

D = 4 + 20π

Hooray! We finally have the position function:

s(t) = -10sin(t) - 3cos(t) - 10t + 4 + 20π

And there you have it! The position function that describes our happy little particle's journey. I hope this explanation made you smile amidst all the calculations!

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.

To find a function that describes the position of the particle, you need to integrate the acceleration function to obtain the velocity function, and then integrate the velocity function to obtain the position function.

1. Integration of acceleration to obtain velocity:
To integrate the function a(t) = 10sin(t) + 3cos(t), we need to know the antiderivatives of sin(t) and cos(t). The antiderivative of sin(t) is -cos(t), and the antiderivative of cos(t) is sin(t).

So, we have:

∫[10sin(t) + 3cos(t)] dt = -10cos(t) + 3sin(t) + C₁

Where C₁ is the constant of integration.

2. Integration of velocity to obtain position:
To integrate the velocity function, -10cos(t) + 3sin(t) + C₁, we integrate each term separately. The antiderivative of -10cos(t) is -10sin(t), and the antiderivative of 3sin(t) is -3cos(t). The constant C₁ integrates to C₂, where C₂ is a new constant of integration.

So, we have:

s(t) = ∫[-10cos(t)] dt + ∫[3sin(t)] dt + C₂
= -10sin(t) - 3cos(t) + C₂

3. Applying the given initial condition to find C₂:
We are given that s(0) = -4. We can substitute t = 0 into the position function to find the value of C₂:

-10sin(0) - 3cos(0) + C₂ = -4
-3 + C₂ = -4
C₂ = -1

Thus, the position function s(t) is:

s(t) = -10sin(t) - 3cos(t) - 1

So, the function describing the position of the particle is s(t) = -10sin(t) - 3cos(t) - 1.