Step-by-step explanation:
Know that
The derivative of position with respect of t is velocity.
Or if you want to say
[tex] \frac{d}{dt} (s(t) = v[/tex]
Or in the form of the definition of the derivative
[tex]s'(t) = \frac{6 (t + h) {}^{3} + 2(t + h) + 9 - (6 {t}^{3} + 2t + 9) }{h} [/tex]
[tex]s'(t) = \frac{6 {t}^{3} + 18t {}^{2}h + 18h {}^{2} t + 6 {h}^{3} + 2t + 2h + 9 - (6 {t}^{3} + 2t + 9)}{h} [/tex]
[tex]s'(t) = \frac{18 {t}^{2}h + 18h {}^{2}t + 6h {}^{3} + 2h }{h} [/tex]
[tex]s'(t) = 18 {t}^{2} + 18th + 6h {}^{2} + 2[/tex]
Let h be a small number, so we can cancel out small things.
[tex]s'(t) = 18 {t}^{2} + 2[/tex]
We know the derivative of position is velocity
[tex]v = 18 {t}^{2} + 2[/tex]
So the velocity at time t is
[tex]v = 18 {t}^{2} + 2[/tex]
Plug in 3 for t to find the velocity
[tex]v(3) =18(3) {}^{2} + 2[/tex]
[tex]v(3) = 164[/tex]
So the velocity at t=3, is 164.
Next, the derivative of velocity is acceleration so
[tex]v'(t) = \frac{18(t + h) {}^{2} + 2 - (18 {t}^{2} + 2)}{h} [/tex]
[tex]v'(t) = \frac{18 {t}^{2} + 36ht + 18 {h}^{2} + 2(- 18 {t}^{2} + 2 ) }{h} [/tex]
[tex]v'(t) = \frac{36ht + 18 {h}^{2} }{h} [/tex]
Divide by h.
[tex]v'(t) = 36t + 18h[/tex]
[tex]v'(t) = 36t[/tex]
The derivative of velocity is acceleration so
[tex]a = 36t[/tex]
The acceleration in time t is
[tex]36t[/tex]
To find acceleration, plug in 3 for t.
[tex]a(3) = 36( 3) = 108[/tex]
