Question

a small frictionless cart, attached to the wall by a spring, is pulled 10 cm from its rest position and released at time t = 0 to roll back and forth for 4 sec. Its position at time t is s=10cos(pi)t.

what is the cart's maximum speed?

Answer 1

at t=0. carts gain maximum speed.
after t=0, speed wll be constant bcs the frictionless surface.

Answer 2

*how do you know? do you find the derivative?

Answer 3

\[s=10\cos(\pi t)\]\[v=-10\pi\sin (\pi t)\]\[a=-10\pi^2\cos (\pi t)=0 \to t=1/2,3/2,5/2,...\]plug that back into v and look at absolute value for speed: \[v_{\max}=\left| -10\pi \sin( {\pi \over2})\right|=10 \pi\]

Answer 4

so it is looking at acceleration, not velocity?

Answer 5

Sorry, I am confused.. why do you want to find values with a = 0?

Answer 6

No, we maximized the velocity function by taking it's derivative and setting it to zero.
Just like for any function f(x) we find max/mins by finding f'(x)=0
v'(t)=0
It just so happens that in physics v'(t)=a(t)
So we got our formula for v(t), took the derivative, and set it equal to zero to find the value of t that maximizes v(t).

Answer 7

another way to see this and get the same result without taking two derivatives is to realize that an oscillating object is moving fastest at it's equilibrium position, i.e when s(t)=0:\[s(t)=10\cos(\pi t)=0 \to t={1\over2},{3\over2},{5\over2},...\]Now find v(t)=s'(t):\[v(t)=s'(t)=-10\pi \sin(\pi t)\]Now plug in one of our critical values of t to find the max and we get the same answer.

This last method is simpler because it requires less work, but you need to know the physics of an simple harmonic oscillator first to know that v max occurs at equilibrium position.

The other method requires no such knowledge, just basic calculus; but it does require you to take an extra derivative to maximize v(t).

We get the same answer either way though, so the physics is consistent :)