How do you find the instantaneous velocity at #t=2# for the position function #s(t) = t^3 +8t^2-t#?

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Answer 1 · 2018-04-13T00:28:53

#43#

The instantaneous velocity is given by #(ds)/dt#.

Since #s(t)=t^3+8t^2-t#, #(ds)/dt=3t^2+16t-1#.

At #t=2#, #[(ds)/dt]_(t=2)=3*2^2+16*2-1=43#.

Answer 2 · 2018-04-13T00:31:50

#43#

We have the position as the function #s(t)=t^3+8t^2-t#.

Velocity is the rate of change of position over time, so its the derivative of the function.

#:.s'(t)=3t^2+16t-1#

So at #t=2#, the velocity is,

#s'(2)=3*2^2+16*2-1#

#=3*4+32-1#

#=12+32-1#

#=44-1#

#=43#