How do you find the instantaneous velocity at t=2t=2 for the position function s(t) = t^3 +8t^2-ts(t)=t3+8t2t?

2 Answers
Apr 13, 2018

4343

Explanation:

The instantaneous velocity is given by (ds)/dtdsdt.

Since s(t)=t^3+8t^2-ts(t)=t3+8t2t, (ds)/dt=3t^2+16t-1dsdt=3t2+16t1.

At t=2t=2, [(ds)/dt]_(t=2)=3*2^2+16*2-1=43[dsdt]t=2=322+1621=43.

Apr 13, 2018

4343

Explanation:

We have the position as the function s(t)=t^3+8t^2-ts(t)=t3+8t2t.

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