A particle moves along a straight line with displacement s(t), velocity v(t), and acceleration a(t). Show that a(t) = v(t) dv/ds.
Well, this all sounds very fancy and scientific, but let's break it down with a touch of humor!
So we have this particle moving along a straight line, right? Let's imagine the particle is on a scenic road trip. Displacement, or s(t), is like the total distance it has traveled so far.
Now, velocity, or v(t), is how fast the particle is zooming along that road. And acceleration, or a(t), is basically how quickly the particle changes its speed.
To connect all these fancy terms, we need a good ol' formula! And that formula is a(t) = v(t) dv/ds.
I like to think of dv/ds as "Rate of Speed Change Per Unit of Road Trip." It basically tells us how much the particle's velocity changes relative to the distance it has traveled.
So this formula is saying that the particle's acceleration is equal to its velocity multiplied by this "Rate of Speed Change Per Unit of Road Trip."
And there you have it! The road trip equation that relates acceleration, velocity, and displacement. Just remember, safety first on those theoretical road trips!
To show that a(t) = v(t) dv/ds, we can start by using the chain rule of differentiation.
Let's differentiate the displacement function s(t) with respect to time, t:
ds/dt = v(t)
Now, let's differentiate both sides of this equation with respect to t:
d^2s/dt^2 = dv/dt
This is the second derivative of s(t) with respect to t, which represents the acceleration a(t):
a(t) = dv/dt
Next, let's rewrite the equation ds/dt = v(t) in terms of differentials:
ds = v(t) dt
Now, we can differentiate both sides of this equation with respect to s:
d(ds) = d(v(t) dt)
1 = v(t) dv/ds (using the chain rule)
Therefore, we have shown that a(t) = v(t) dv/ds.