The acceleration of a certain rocket is given by a=bt where b is positive cnstant, find the position function Xo and Vo at t=o?
Problem reposted and answered here:
http://www.jiskha.com/display.cgi?id=1255001673
To find the position function Xo and the velocity function Vo at t = 0, we need to integrate the given acceleration function with respect to time.
Given: a = bt
Let's integrate the acceleration function to obtain the velocity function:
∫ a dt = ∫ bt dt
Using the power rule of integration, the integral of bt with respect to t is (1/2)bt^2:
∫ a dt = (1/2)bt^2 + C1
Where C1 is the constant of integration.
Now, we have: Vo = ∫ a dt = (1/2)bt^2 + C1
To find the constant C1, we have to evaluate the velocity function Vo at t = 0:
Vo(0) = (1/2)b(0)^2 + C1
= C1
Hence, the constant C1 is equal to the initial velocity at t = 0, which we'll denote as Vo(0):
C1 = Vo(0)
Therefore, the velocity function at t = 0 is:
Vo = (1/2)bt^2 + Vo(0)
Next, let's find the position function Xo by integrating the velocity function:
∫ Vo dt = ∫ [(1/2)bt^2 + Vo(0)] dt
Using the power rule of integration, the integral of (1/2)bt^2 with respect to t is (1/6)bt^3:
∫ Vo dt = (1/6)bt^3 + Vo(0)t + C2
Where C2 is the constant of integration.
Now, we have: Xo = ∫ Vo dt = (1/6)bt^3 + Vo(0)t + C2
To find the constant C2, we have to evaluate the position function Xo at t = 0:
Xo(0) = (1/6)b(0)^3 + Vo(0)(0) + C2
= C2
Hence, the constant C2 is equal to the initial position at t = 0, which we'll denote as Xo(0):
C2 = Xo(0)
Therefore, the position function at t = 0 is:
Xo = (1/6)bt^3 + Vo(0)t + Xo(0)
So, the position function Xo and the velocity function Vo at t = 0 are given by:
Xo = (1/6)bt^3 + Vo(0)t + Xo(0)
Vo = (1/2)bt^2 + Vo(0)