An object moves along a coordinate line with acceleration a(t)=(t+2)^3 units per second per second.
a). The initial velocity is 9 units per second.
The velocity function is v(t) =
b).The initial position is 2 units to the right of the origin.
The position function is x(t) =
a). The initial velocity is 9 units per second.
The velocity function can be found by integrating the acceleration function:
v(t) = ∫[a(t)] dt
= ∫[(t+2)^3] dt
= [(1/4)*(t+2)^4] + C
Now, we need to find the value of C. Since the initial velocity is 9 units per second when t = 0, we can substitute these values into the velocity function and solve for C:
9 = [(1/4)*(0+2)^4] + C
9 = (1/4)*(2^4) + C
9 = (1/4)*16 + C
9 = 4 + C
C = 9 - 4
C = 5
Therefore, the velocity function is:
v(t) = [(1/4)*(t+2)^4] + 5
b). The initial position is 2 units to the right of the origin.
The position function can be found by integrating the velocity function:
x(t) = ∫[v(t)] dt
= ∫[[(1/4)*(t+2)^4] + 5] dt
= [(1/20)*(t+2)^5] + 5t + D
Now, we need to find the value of D. Since the initial position is 2 units to the right of the origin when t = 0, we can substitute these values into the position function and solve for D:
2 = [(1/20)*(0+2)^5] + 5*0 + D
2 = (1/20)*(2^5) + D
2 = (1/20)*32 + D
2 = (32/20) + D
2 = 8/5 + D
D = 2 - 8/5
D = 2/5
Therefore, the position function is:
x(t) = [(1/20)*(t+2)^5] + 5t + 2/5
a) To find the velocity function, we need to take the antiderivative of the acceleration function.
Let's denote the variable of integration as t.
So, integrating the given acceleration function a(t) = (t+2)^3 with respect to t, we get:
v(t) = ∫ a(t) dt = ∫ (t+2)^3 dt
Applying the power rule of integration, we have:
v(t) = (1/4)(t+2)^4 + C
Now, we need to find the constant term C. Since the initial velocity is given as 9 units per second, we can substitute the initial conditions to solve for C.
When t = 0, v(t) = 9 units per second.
So, plugging in these values and solving for C:
9 = (1/4)(0+2)^4 + C
9 = (1/4)(2)^4 + C
9 = (1/4)(16) + C
9 = 4 + C
C = 9 - 4
C = 5
Therefore, the velocity function is given by:
v(t) = (1/4)(t+2)^4 + 5
b) To find the position function, we need to take the antiderivative of the velocity function.
Again, let's denote the variable of integration as t.
So, integrating the velocity function v(t) = (1/4)(t+2)^4 + 5 with respect to t, we get:
x(t) = ∫ v(t) dt = ∫ [(1/4)(t+2)^4 + 5] dt
Applying the power rule of integration, we have:
x(t) = (1/20)(t+2)^5 + 5t + C'
Now, we need to find the constant term C'. Since the initial position is given as 2 units to the right of the origin, we can substitute the initial conditions to solve for C'.
When t = 0, x(t) = 2 units to the right of the origin.
So, plugging in these values and solving for C':
2 = (1/20)(0+2)^5 + 5(0) + C'
2 = (1/20)(2)^5 + 0 + C'
2 = (1/20)(32) + C'
2 = (32/20) + C'
40/20 = 32/20 + C'
1 = C'
Therefore, the position function is given by:
x(t) = (1/20)(t+2)^5 + 5t + 1
a) To find the velocity function, we will integrate the acceleration function with respect to time.
The acceleration function is given as a(t) = (t+2)^3 units per second per second.
To integrate this, we use the power rule for integration, which states that when integrating a function of the form x^n, where n is a constant, the result is (1/(n+1)) * x^(n+1) + C, where C is the constant of integration.
In this case, if we integrate (t+2)^3 with respect to t, we get:
∫(t+2)^3 dt = (1/4) * (t+2)^4 + C
Now, since the initial velocity is given as 9 units per second, we can use this information to find the constant of integration, C.
When t = 0 (initial time), v(t) = 9. Substituting this into the velocity function, we get:
v(0) = (1/4) * (0+2)^4 + C
9 = (1/4) * 2^4 + C
9 = (1/4) * 16 + C
9 = 4 + C
C = 9 - 4
C = 5
Thus, the velocity function is:
v(t) = (1/4) * (t+2)^4 + 5
b) To find the position function, we will integrate the velocity function with respect to time.
The velocity function is given as v(t) = (1/4) * (t+2)^4 + 5 units per second.
Integrating this, we get:
∫[(1/4) * (t+2)^4 + 5] dt = (1/20) * (t+2)^5 + 5t + C
Since the initial position is given as 2 units to the right of the origin, we can use this information to find the constant of integration, C.
When t = 0 (initial time), x(t) = 2. Substituting this into the position function, we get:
x(0) = (1/20) * (0+2)^5 + 5 * 0 + C
2 = (1/20) * 2^5 + C
2 = (1/20) * 32 + C
2 = 8/5 + C
C = 2 - 8/5
C = 2/5
Thus, the position function is:
x(t) = (1/20) * (t+2)^5 + 5t + 2/5
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v = 1/3 (t+2)^2 + c1
x = 1/6 (t+2) + c1*t + c2
now plug in the given conditions when t=0 to find c1 and c2