For positive constants k and g, the velocity, v, of a particle of mass m at time t is given by
At what rate is the velocity is changing at time 0? At t=1? What do your answers tell you about the motion?
April 1, 2011 2:49am UTC
rate = g
rate = g*(e^(-k/m))
November 28, 2021 8:23pm UTC
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!
September 23, 2023 11:48pm UTC