Answer:
The instantaneous velocity at t = 1 will be:
- [tex]\frac{d}{dt}\left(-8-9t\right)=-9[/tex]
Step-by-step explanation:
Given the position of an object at a time [tex]t[/tex]
[tex]s\left(t\right)=-8-9t[/tex]
As we know that determining the derivative of the position function with respect to time t would give us the instantaneous velocity, so
[tex]\frac{ds}{dt}=\frac{d}{dt}\left(-8-9t\right)[/tex]
Applying the sum/difference rule:
[tex]\left(f\pm g\right)'=f\:'\pm g'[/tex]
[tex]\frac{ds}{dt}=-\frac{d}{dt}\left(8\right)-\frac{d}{dt}\left(9t\right)[/tex]
as
[tex]\frac{d}{dt}\left(8\right)=0[/tex] ∵ [tex]\frac{d}{dx}\left(a\right)=0[/tex]
and
[tex]\frac{d}{dt}\left(9t\right)=9[/tex] ∵ [tex]\mathrm{Take\:the\:constant\:out}:\quad \left(a\cdot f\right)'=a\cdot f\:'[/tex] and [tex]\frac{d}{dt}\left(t\right)=1[/tex]
so the expression becomes
[tex]\frac{ds}{dt}=-\frac{d}{dt}\left(8\right)-\frac{d}{dt}\left(9t\right)[/tex]
[tex]=-0-9[/tex]
[tex]=-9[/tex]
As the derivative is constant.
Therefore, the instantaneous velocity at t = 1 will be:
- [tex]\frac{d}{dt}\left(-8-9t\right)=-9[/tex]
