Respuesta :
Answer:
The velocity function is [tex]v(t)=10t+5[/tex].
The acceleration function is [tex]a(t)=10[/tex].
When t = 44, the velocity is [tex]v(44)=445 \:\frac{ft}{s}[/tex].
When t = 44, the acceleration is [tex]a(44)=10\: \frac{ft}{s^2}[/tex].
Step-by-step explanation:
We know that the position function is given by
[tex]s(t)=5t^2+5t[/tex]
Velocity is defined as the rate of change of position or the rate of displacement. If you take the derivative of the position function you get the instantaneous velocity function.
[tex]v(t)=\frac{ds}{dt}[/tex]
Acceleration is defined as the rate of change of velocity. If you take the derivative of the instantaneous velocity function you get the instantaneous acceleration function.
[tex]a(t)=\frac{dv}{dt}[/tex]
The instantaneous velocity function is given by
[tex]v(t)=\frac{d}{dt} s(t)=\frac{d}{dt}(5t^2+5t)\\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\v(t)=\frac{d}{dt}\left(5t^2\right)+\frac{d}{dt}\left(5t\right)\\\\\mathrm{Apply\:the\:Power\:Rule}:\quad \frac{d}{dx}\left(x^a\right)=a\cdot x^{a-1}\\\\v(t)=10t+\frac{d}{dt}\left(5t\right)\\\\\mathrm{Apply\:the\:common\:derivative}:\quad \frac{d}{dt}\left(t\right)=1\\\\v(t)=10t+5[/tex]
The instantaneous acceleration function is given by
[tex]a(t)=\frac{dv}{dt} =\frac{d}{dt}(10t+5)\\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\a(t)=\frac{d}{dt}\left(10t\right)+\frac{d}{dt}\left(5\right)\\\\a(t)=10[/tex]
To find the velocity and acceleration when t = 44, we substitute this value into the velocity and acceleration functions
[tex]v(44)=10(44)+5\\v(44)=445 \:\frac{ft}{s}[/tex]
[tex]a(44)=10\: \frac{ft}{s^2}[/tex]
