1 pt) For positive constants k and g, the velocity, v, of a particle of mass m at time t is given by
v=mg/k(1−e−kt/m).
At what rate is the velocity is changing at time 0? At t=8? What do your answers tell you about the motion?
At what rate is the velocity changing at time 0, time 8?
To find the rate at which the velocity is changing, we need to calculate the derivative of the velocity equation with respect to time, t.
Let's start by finding the derivative of the velocity equation at time t=0. We will use the chain rule since we have a composite function: v(t) = mg/k(1−e^(-kt/m)).
First, we differentiate the outer function, which is mg/k, treating m, g, and k as constants. The derivative of a constant times a function is simply the constant multiplied by the derivative of the function.
So, d/dt (mg/k) = 0.
Next, we differentiate the inner function, which is (1-e^(-kt/m)). Remember, the derivative of e^x is e^x.
d/dt (1 − e^(-kt/m)) = 0 - (1/m)e^(-kt/m) * (−k/m).
Simplifying, we have d/dt (1 − e^(-kt/m)) = (k/m^2)e^(-kt/m).
Since the outer function derivative is 0, the overall derivative is 0.
So, at time t=0, the rate of change of velocity is 0.
Now let's find the rate of change of velocity at t=8.
d/dt (mg/k(1 − e^(-kt/m))) = (mg/k) * (k/m^2)e^(-kt/m).
Notice that as t gets larger, the exponential term e^(-kt/m) approaches zero. Therefore, at t=8, the rate of change of velocity is also 0.
The fact that the rate of change of velocity is 0 at both t=0 and t=8 tells us that the particle has reached a steady state. This means that the particle is no longer accelerating or decelerating, but moving at a constant velocity.
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