Acceleration with respect to position

A particle travels along a straight line with a velocity of v=(21−0.05s²) m/s, where s is in meters.

Determine the acceleration of the particle at s = 16 m . Express your answer to three significant figures and include the appropriate units.

We need to find a formula that gives us acceleration in terms of position. Easy!

v = ds/dt

a = dv/dt

Here is the fun part!

You could place a form of one in the equation! Like this form of one: ds/ds

a = dv/dt * ds/ds

Group them!

a = (dv*ds)/(ds/dt)

Now separate them in a new way

a = dv/ds * ds/dt

We know that v = ds/dt

a = dv/ds * v

And we were given v = 21-.05s²

a = dv/ds * 21 - .05s²

and dv/ds is simply the derivative of the given equation with respect to s

a = -.1s (21-.05s²

Plug in 16,

a(16)=-13.1 m/s²