find the position function s(t) given acceleration a(t)=3t, if v(2)=0, and s(2)=1.
the answer is s(t)=(t^3/2)-6t +9 but I don't know how to solve this.
Can someone help me with this?
Thank you!
sure you do.
a(t) = 3t
v(t) = 3/2 t^2 + c
but, v(2)=0, so
3/2(4)+c = 0
c = -6, so
v(t) = 3/2 t^2 - 6
s(t) = 1/2 t^3 - 6t + c
but, s(2)=1, so
1/2 (8) - 6(2) + c = 1
c = 9
s(t) = 1/2 t^3 - 6t + 9
To find the position function s(t) given the acceleration a(t) = 3t, we need to integrate the acceleration function twice.
First, we integrate the acceleration function to find the velocity function v(t):
∫a(t) dt = ∫3t dt = (3/2)t^2 + C1
Using the given initial condition v(2) = 0, we can solve for the constant C1:
v(2) = (3/2)(2)^2 + C1 = 0
6 + C1 = 0
C1 = -6
So, the velocity function becomes:
v(t) = (3/2)t^2 - 6
Now, we integrate the velocity function v(t) to find the position function s(t):
∫v(t) dt = ∫[(3/2)t^2 - 6] dt = (1/2)t^3 - 6t + C2
Using the given initial condition s(2) = 1, we can solve for the constant C2:
s(2) = (1/2)(2)^3 - 6(2) + C2 = 1
4 - 12 + C2 = 1
C2 = 9
So, the position function becomes:
s(t) = (1/2)t^3 - 6t + 9
Hence, the position function is s(t) = (1/2)t^3 - 6t + 9.
To find the position function, we can integrate the acceleration function twice.
Step 1: Integrate acceleration function to find velocity function.
Given: a(t) = 3t
To find the velocity function v(t), integrate the acceleration function with respect to time:
∫a(t) dt = ∫3t dt
Integrating 3t with respect to t gives us: 3/2 * t^2 + C1
C1 is the constant of integration.
Step 2: Use initial velocity condition to find the value of C1.
Given: v(2) = 0
Substitute t = 2 into the velocity function and solve for C1:
0 = 3/2 * 2^2 + C1
0 = 6 + C1
C1 = -6
So, the velocity function v(t) is:
v(t) = 3/2 * t^2 - 6
Step 3: Integrate the velocity function to find the position function.
To find the position function s(t), integrate the velocity function with respect to time:
∫v(t) dt = ∫(3/2 * t^2 - 6) dt
Integrating (3/2 * t^2 - 6) with respect to t gives us: 1/2 * t^3 - 6t + C2
C2 is the constant of integration.
Step 4: Use the initial position condition to find the value of C2.
Given: s(2) = 1
Substitute t = 2 into the position function and solve for C2:
1 = 1/2 * 2^3 - 6 * 2 + C2
1 = 4 - 12 + C2
1 = -8 + C2
C2 = 9
So, the position function s(t) is:
s(t) = 1/2 * t^3 - 6t + 9
Therefore, the answer is s(t) = (t^3/2) - 6t + 9.