Final answer:
To solve for the particle's position, velocity, and acceleration as functions of time, we integrate the acceleration function first to find the velocity and then the position. We find that the velocity function is v(t) = 28e^{-3t} m/s, the position function is s(t) = -(28/3)e^{-3t} + (28/3)t m, and the acceleration is a(t) = -84e^{-3t} m/s².
Step-by-step explanation:
To determine the particle's position, velocity, and acceleration as functions of time when given an acceleration that is a function of the velocity, we first integrate the acceleration to find the velocity, and then integrate the velocity to find the position. Let's solve the given problem where the acceleration a(t) is equal to -3 times the velocity v(t).
Given:
- Acceleration a(t) = -3v
- Initial velocity v(0) = 28 m/s
- Initial position s(0) = 0
First, we separate variables in the acceleration function to find the velocity function v(t):
dv/dt = -3v
dv/v = -3 dt
Integrating both sides, we get:
ln(v) = -3t + C
To determine the value of C, we use the initial condition v(0) = 28 m/s:
ln(28) = C
Now we solve for v:
v(t) = e^{-3t + ln(28)}
v(t) = 28e^{-3t}
Next, we integrate the velocity function to find the position function s(t):
ds/dt = 28e^{-3t}
Integrating:
s(t) = -(28/3)e^{-3t} + Ct
Using the initial condition s(0) = 0, we can find C:
s(0) = -(28/3)e^{0} + C(0) = 0
C = 28/3
Thus the position function is:
s(t) = -(28/3)e^{-3t} + (28/3)t
Finally, using the original acceleration function, we have:
a(t) = -3 * 28e^{-3t}
a(t) = -84e^{-3t}
To summarize, the functions are:
- Velocity: v(t) = 28e^{-3t} m/s
- Position: s(t) = -(28/3)e^{-3t} + (28/3)t m
- Acceleration: a(t) = -84e^{-3t} m/s²