Can you please check my work.

A particle is moving with the given data. Find the position of the particle.
a(t) = cos(t) + sin(t)
s(0) = 2
v(0) = 6

a(t) = cos(t) + sin(t)
v(t) = sin(t) - cos(t) + C
s(t) = -cos(t) - sin(t) + Cx + D

6 = v(0) = sin(0) -cos(0) + C
C=7

2= s(0) = -cos(0) - sin(0) + 7 (0) + D
D= 3

s(t) = -cos(t) - sin(t) + 7t + 3

Well, well, well! Let's see if your work matches my expectations.

First off, you found the velocity equation v(t) correctly. Great job!

Next, when you integrated v(t), you included two constants C and D. However, you made a small mistake in your s(t) equation. The term "Cx" should actually be just "C," since there is no "t" term attached to it.

So, the correct equation should be:

s(t) = -cos(t) - sin(t) + 7 + D

Now, let's substitute the initial conditions to find the values of C and D.

You mentioned that v(0) = 6. Let's put that into the v(t) equation:

6 = sin(0) - cos(0) + C

And correct me if I'm wrong, but sin(0) = 0 and cos(0) = 1. So we have:

6 = 0 - 1 + C

This simplifies to:

6 = -1 + C

Adding 1 to both sides, we get:

7 = C

Now let's move on to s(0) = 2:

2 = -cos(0) - sin(0) + 7 + D

Since cos(0) = 1 and sin(0) = 0, we have:

2 = -1 + 0 + 7 + D

This simplifies to:

2 = 6 + D

Subtracting 6 from both sides, we get:

-4 = D

Putting it all together, the correct equation for the position of the particle is:

s(t) = -cos(t) - sin(t) + 7 + (-4)

Or, simplifying further:

s(t) = -cos(t) - sin(t) + 3

Keep up the good work, my friend!

To find the position of the particle, we can integrate the velocity function with respect to time to get the position function.

First, let's integrate the acceleration function to get the velocity function:

∫[a(t)] dt = ∫[(cos(t) + sin(t))] dt

The integral of cos(t) is sin(t), and the integral of sin(t) is -cos(t).

Therefore, v(t) = sin(t) - cos(t) + C, where C is the constant of integration.

Since we know that v(0) = 6, we can substitute t = 0 into the velocity function to solve for C:

6 = sin(0) - cos(0) + C
6 = 0 - 1 + C
C = 7

Now that we have the value of C, we can find the position function by integrating the velocity function:

∫[v(t)] dt = ∫[(sin(t) - cos(t) + 7)] dt

The integral of sin(t) is -cos(t), the integral of -cos(t) is -sin(t), and the integral of 7 is 7t.

Therefore, s(t) = -cos(t) - sin(t) + 7t + D, where D is the constant of integration.

Since we know that s(0) = 2, we can substitute t = 0 into the position function to solve for D:

2 = -cos(0) - sin(0) + 7(0) + D
2 = -1 - 0 + 0 + D
D = 3

Therefore, the position function is:

s(t) = -cos(t) - sin(t) + 7t + 3

To check your work, we need to verify that the equations you derived correctly describe the motion of the particle.

Let's start with the acceleration function:

a(t) = cos(t) + sin(t)

To find the velocity function, integrate the acceleration function with respect to time:

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

Here, C is the constant of integration.

Next, we integrate the velocity function to find the position function:

∫v(t) dt = ∫(sin(t) - cos(t) + C) dt
s(t) = -cos(t) - sin(t) + Ct + D

Here, C and D are constants of integration.

Now, let's check the given conditions:

s(0) = 2:
Plugging in t = 0 into the position function, we have:
s(0) = -cos(0) - sin(0) + C(0) + D
s(0) = -1 + 0 + 0 + D
s(0) = -1 + D

Since s(0) is given as 2, we can solve for D:
2 = -1 + D
D = 3

v(0) = 6:
Plugging in t = 0 into the velocity function, we have:
v(0) = sin(0) - cos(0) + C
v(0) = 0 + 1 + C
v(0) = 1 + C

Since v(0) is given as 6, we can solve for C:
6 = 1 + C
C = 5

Now we can substitute the values of C and D into the position function to get the final expression for s(t):

s(t) = -cos(t) - sin(t) + 5t + 3

Therefore, your final expression for the position of the particle is:

s(t) = -cos(t) - sin(t) + 5t + 3.

All correct.

We can check backwards:
assume given:
s(t) = -cos(t) - sin(t) + 7t + 3

v(t)=s'(t)=sin(t)-cos(t)+7
a(t)=v'(t)=cos(t)+sin(t)
v(0)=sin(0)-cos(0)+7=0-1+7=6
s(0)=-cos(0)-sin(0) + 7(0) + 3 =-1+0+0+3=2
All correct!