Question

in an example it was given -
v^2=-10s+2500
2vdt/dt=-10ds/dt
please explain the second step.

Answer 1

there's a typo in this (dt/dt) but we can do it anyway

you start with
\[v^2=-10s+2500 \]

you are now going to differentiate with respect to t
this requires *implicit* differentiation
\[\frac{d \ (v^2)}{dt} = 2 \ v \ \frac{dv}{dt}\]
\[\frac{d \ (-10s)}{dt} = -10 \ \frac{ds}{dt}\]

Answer 2

thanks @IrishBoy123
:)

Answer 3

Picklein typo in that book got me here