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

Answer 1

Elijah Abram

#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#.

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Answer 2

Christopher Agresta

#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#

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Answer 3

Emily Ammerman

To find the instantaneous velocity at ( t = 2 ) for the position function ( s(t) = t^3 + 8t^2 - t ), you need to find the derivative of the position function ( s(t) ) with respect to time ( t ), which gives the velocity function ( v(t) ). Then, evaluate ( v(t) ) at ( t = 2 ) to find the instantaneous velocity at that point.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.