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!
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
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!
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!
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
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.