To calculate the position function x(t) for the particle undergoing acceleration given by a = -bv, you can integrate the equation of motion.
Let's start by solving the differential equation. We have:
a = -bv
Since acceleration is the derivative of velocity with respect to time, we can rewrite the equation as:
dv/dt = -bv
To solve this differential equation, we can separate variables:
dv/v = -b dt
Integrating both sides, we get:
∫(dv/v) = -∫b dt
ln|v| = -bt + C
Where C is the constant of integration. Taking the exponential of both sides:
|v| = e^(-bt+C)
Since the absolute value is used, we can remove the absolute value sign by introducing a positive constant D:
v = De^(-bt)
Now, we need to solve for the position x(t). The velocity v(t) is the derivative of the position x(t) with respect to time:
v = dx/dt
To find x(t), we integrate the velocity expression:
∫v dt = ∫De^(-bt) dt
x(t) = -D/b * e^(-bt) + C'
Where C' is another constant of integration. We can rewrite this equation using initial conditions. Let's assume at t = 0, the velocity is v0, and the position is x0:
v(0) = v0
x(0) = x0
Substituting these initial conditions into the equation:
v(0) = D * e^(-b * 0)
v0 = D
x(0) = -D/b * e^(-b * 0) + C'
x0 = -D/b + C'
C' = x0 + D/b
Substituting the value of D obtained from the initial velocity condition into C':
C' = x0 + v0/b
Finally, we can rewrite the position function x(t) in terms of x0, v0, and the constant b:
x(t) = -v0/b * e^(-bt) + x0 + v0/b * e^(-b * 0)
x(t) = x0 + (v0/b) * (1 - e^(-bt))
So, the answer is x(t) = x0 + (v0/b) * (1 - e^(-bt)).