Find the position function s(t) from the given acceleration function and initial values
a (t) = 10-3t^2 ; if v(0) = 3 and s(0) =2
Well, well, well! Looks like we have some physics in our hands. Don't worry, I'll do my best to clown around and help you out with the position function.
To find the position function, we need to integrate the acceleration function twice. Let's break it down step by step, shall we?
First, let's integrate the acceleration function with respect to time to find the velocity function:
v(t) = ∫ (10 - 3t^2) dt
Integrating 10 with respect to t gives us 10t. And integrating -3t^2 with respect to t gives us -t^3. So the velocity function becomes:
v(t) = 10t - t^3 + C₁
Now, we can use the initial condition v(0) = 3 to solve for C₁:
3 = 10(0) - (0)^3 + C₁
3 = 0 + 0 + C₁
C₁ = 3
So the velocity function becomes:
v(t) = 10t - t^3 + 3
Now, let's integrate the velocity function with respect to time to find the position function:
s(t) = ∫ (10t - t^3 + 3) dt
Integrating 10t with respect to t gives us 5t^2. Integrating -t^3 with respect to t gives us -t^4/4. And integrating 3 with respect to t gives us 3t. So the position function becomes:
s(t) = 5t^2 - t^4/4 + 3t + C₂
Finally, we can use the initial condition s(0) = 2 to solve for C₂:
2 = 5(0)^2 - (0)^4/4 + 3(0) + C₂
2 = 0 - 0 + 0 + C₂
C₂ = 2
So the final position function is:
s(t) = 5t^2 - t^4/4 + 3t + 2
And there you have it! The position function to clown around with. Have fun with your physics antics and remember, always stay in motion, but never out of laughter!
To find the position function s(t), we need to integrate the acceleration function a(t) twice with respect to time (t).
Given:
a(t) = 10 - 3t^2
v(0) = 3 (initial velocity)
s(0) = 2 (initial position)
Step 1: Find the velocity function v(t)
To find v(t), integrate a(t) with respect to t:
∫ a(t) dt = ∫ (10 - 3t^2) dt
The integral of 10 with respect to t is 10t, and the integral of -3t^2 with respect to t is -t^3.
So, v(t) = 10t - t^3/3 + C, where C is the constant of integration.
Step 2: Use the initial velocity condition to find the value of C
Given that v(0) = 3, substitute t = 0 and v(t) = 3 into the velocity function:
v(0) = 10(0) - (0)^3/3 + C = 0 - 0 + C = C
Therefore, the constant of integration (C) is 3.
So, the velocity function becomes: v(t) = 10t - t^3/3 + 3.
Step 3: Find the position function s(t)
To find s(t), integrate v(t) with respect to t:
∫ v(t) dt = ∫ (10t - t^3/3 + 3) dt
The integral of 10t with respect to t is 5t^2, the integral of -t^3/3 with respect to t is -t^4/12, and the integral of 3 with respect to t is 3t.
So, s(t) = 5t^2 - t^4/12 + 3t + C2, where C2 is the constant of integration.
Step 4: Use the initial position condition to find the value of C2
Given that s(0) = 2, substitute t = 0 and s(t) = 2 into the position function:
s(0) = 5(0)^2 - (0)^4/12 + 3(0) + C2 = 0 - 0 + 0 + C2 = C2
Therefore, the constant of integration (C2) is 2.
So, the position function becomes: s(t) = 5t^2 - t^4/12 + 3t + 2.
The position function s(t) is 5t^2 - t^4/12 + 3t + 2.
To find the position function s(t) from the given acceleration function a(t), we will integrate the acceleration function twice. Given that v(0) = 3 and s(0) = 2, we will use these initial conditions to find the constant of integration.
Step 1: Finding the velocity function v(t)
Since acceleration is the derivative of velocity, we will integrate the acceleration function a(t) to find the velocity function v(t).
Integrating a(t) = 10 - 3t^2 with respect to t:
∫ a(t) dt = ∫ (10 - 3t^2) dt
Using the power rule of integration, we get:
v(t) = 10t - t^3 + C1
Step 2: Finding the position function s(t)
Since velocity is the derivative of position, we will integrate the velocity function v(t) to find the position function s(t).
Integrating v(t) = 10t - t^3 + C1 with respect to t:
∫ v(t) dt = ∫ (10t - t^3 + C1) dt
Using the power rule of integration, we get:
s(t) = 5t^2 - 1/4 t^4 + C1t + C2
Step 3: Applying the initial conditions
Using the initial conditions v(0) = 3 and s(0) = 2, we can find the values of the integration constants C1 and C2.
v(0) = 3:
Replacing t with 0 in v(t) = 10t - t^3 + C1:
v(0) = 10(0) - (0)^3 + C1
3 = C1
s(0) = 2:
Replacing t with 0 in s(t) = 5t^2 - (1/4) t^4 + C1t + C2:
s(0) = 5(0)^2 - (1/4)(0)^4 + C1(0) + C2
2 = C2
Finally, substituting C1 = 3 and C2 = 2 back into the position function s(t):
s(t) = 5t^2 - 1/4 t^4 + 3t + 2
Therefore, the position function is s(t) = 5t^2 - 1/4 t^4 + 3t + 2.
a = 10-3t^2
v = 10t-t^3 + C
v(0) = 3, so C=3, and
v = 3-10t-t^3
s = 3t - 5t^2 - 1/4 t^4 + C
s(0) = 2, so
s = 2 + 3t - 5t^2 - 1/4 t^4