For positive constants k and g, the velocity, v, of a particle of mass m at time t is given by

Question

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=1? What do your answers tell you about the motion?

Answer 1

t=0

rate = g
t=1
rate = g*(e^(-k/m))

Answer 2

Well, isn't this a mathematical rollercoaster! Let's break it down, shall we?

To find the rate at which velocity is changing at time 0, we need to find dv/dt at t = 0. Now, if you take a look at the equation, you'll notice that when t = 0, the exponential term becomes 0. So, all you're left with is v = mg/k.

But, wait! Don't get too excited just yet! This only tells us the initial velocity at t = 0, not the rate at which it's changing.

To find the rate at which velocity is changing at t = 0, we need to find dv/dt. Taking the derivative of the velocity equation with respect to t gives us:

dv/dt = mg/k * (k/m * e^(-kt/m))

Now, substituting t = 0 into this equation, the exponential term e^(-kt/m) becomes 1, giving us:

dv/dt = mg/k * (k/m * 1) = g

Haha! The rate at which velocity is changing at time 0 is simply g, the constant! That tells us that at the beginning of the motion, the velocity is changing at the rate of gravity, which is quite a "down-to-Earth" observation, don't you think?

Now, let's move on to t = 1.

When t = 1, we can use the same logic. The exponential term e^(-kt/m) at t = 1 is equal to e^(-k/m), and if we plug it into the derivative equation, we get:

dv/dt = mg/k * (k/m * e^(-k/m))

Now, evaluate this equation:

dv/dt = mg/k * (k/m * e^(-k/m)) ≈ 0.632 * g

Voilà! At t = 1, the rate at which the velocity is changing is approximately 0.632 times g. This tells us that the rate of change of velocity is decreasing, which implies that the particle is gradually slowing down. It's like it's taking a leisurely stroll through the realm of motion.

I hope my humorous explanation wasn't too "rate" for you! Mathematics can be a rollercoaster of emotions, but it's all part of the fun!

Answer 3

To find the rate at which the velocity is changing at time t=0, we need to calculate the derivative of the velocity function with respect to time (t).

Given: v = mg/k(1 − e^(-kt/m))

Taking the derivative of v with respect to t:
dv/dt = mg/k * ((0 - (-k/m)) * e^(-kt/m))

Simplifying:
dv/dt = mg/k * (k/m) * e^(-kt/m)

Simplifying further:
dv/dt = g * e^0

Since e^0 equals 1, the expression becomes:
dv/dt = g

Therefore, the rate at which the velocity is changing at time t=0 is g.

Now, let's find the rate at which the velocity is changing at t=1. Substitute t=1 into the velocity function:

v(1) = mg/k(1 − e^(-k/m))

To find the rate of change at t=1, we need to calculate dv/dt at t=1.

Taking the derivative of v with respect to t and evaluating at t=1:
dv/dt = mg/k * ((0 - (-k/m)) * e^(-k/m))

Simplifying:
dv/dt = mg/k * (k/m) * e^(-k/m)

Substituting t=1:
dv/dt = g * e^(-k/m)

The rate at which the velocity is changing at t=1 is given by g * e^(-k/m).

These answers tell us about the motion of the particle:

1. At time t=0, the rate at which the velocity is changing is equal to g. This means that at the initial moment, the particle is experiencing a constant rate of change in velocity of magnitude g.

2. At time t=1, the rate of change in velocity is given by g * e^(-k/m). The exponential term e^(-k/m) indicates that the rate of change in velocity decreases over time. Therefore, at t=1, the particle is still experiencing a change in velocity, but the rate of that change is decreasing.

Answer 4

To find the rate of change of velocity at a specific time, we need to differentiate the velocity function with respect to time (t). Let's start by differentiating the given velocity function in terms of t:

v = mg/k(1 − e^(-kt/m))

To find the rate of change of velocity at time 0, we differentiate the function and substitute t = 0:

dv/dt = (mg/k) * (0 - (-k/m) * e^(-kt/m))
= (mg/k) * (k/m) * e^0
= (mg/k) * (k/m)
= g

So, at time t = 0, the rate of change of velocity is equal to g, which is a positive constant. This indicates that the velocity is changing at a constant rate.

Now, let's find the rate of change of velocity at time t = 1:

dv/dt = (mg/k) * (1 - (-k/m) * e^(-k(1)/m))
= (mg/k) * (1 + k/m * e^(-k/m))

Since t = 1, we substitute it into the equation. The result will depend on the specific values of k, g, and m.

In general, the rate of change of velocity at time t = 1 will depend on the values of the constants k, g, and m. It can be positive, negative, or zero, indicating acceleration, deceleration, or no change in the velocity, respectively.