Given the position function [tex]\( s = f(t) = 5t^3 + 4t + 9 \)[/tex]:
(a) To find the velocity [tex]\( v(t) \)[/tex], we need to take the derivative of the position function with respect to time [tex]\( t \)[/tex]. The velocity function is given by:
[tex]\[
v(t) = \frac{ds}{dt} = \frac{d}{dt}(5t^3 + 4t + 9)
\][/tex]
Using the power rule for differentiation ([tex]\( \frac{d}{dt} (t^n) = n t^{n-1} \)[/tex]):
[tex]\[
v(t) = 3 \cdot 5t^2 + 4 \cdot 1 + 0 = 15t^2 + 4
\][/tex]
Therefore, the velocity at time [tex]\( t \)[/tex] is:
[tex]\[
v(t) = 15t^2 + 4 \quad \frac{m}{s}
\][/tex]
(b) To find the velocity at [tex]\( t = 3 \)[/tex] seconds, we substitute [tex]\( t = 3 \)[/tex] into the velocity function:
[tex]\[
v(3) = 15(3)^2 + 4 = 15 \cdot 9 + 4 = 135 + 4 = 139
\][/tex]
Hence, the velocity at [tex]\( t = 3 \)[/tex] seconds is:
[tex]\[
139 \quad \frac{m}{s}
\][/tex]
(c) To find the acceleration [tex]\( a(t) \)[/tex], we need to take the derivative of the velocity function with respect to time [tex]\( t \)[/tex]. The acceleration function is given by:
[tex]\[
a(t) = \frac{dv}{dt} = \frac{d}{dt}(15t^2 + 4)
\][/tex]
Again, using the power rule for differentiation:
[tex]\[
a(t) = 2 \cdot 15t + 0 = 30t
\][/tex]
Therefore, the acceleration at time [tex]\( t \)[/tex] is:
[tex]\[
a(t) = 30t \quad \frac{m}{s^2}
\][/tex]
(d) To find the acceleration at [tex]\( t = 3 \)[/tex] seconds, we substitute [tex]\( t = 3 \)[/tex] into the acceleration function:
[tex]\[
a(3) = 30 \cdot 3 = 90
\][/tex]
Hence, the acceleration at [tex]\( t = 3 \)[/tex] seconds is:
[tex]\[
90 \quad \frac{m}{s^2}
\][/tex]
In summary:
(a) [tex]\( v(t) = 15t^2 + 4 \ \frac{m}{s} \)[/tex]
(b) [tex]\( v(3) = 139 \ \frac{m}{s} \)[/tex]
(c) [tex]\( a(t) = 30t \ \frac{m}{s^2} \)[/tex]
(d) [tex]\( a(3) = 90 \ \frac{m}{s^2} \)[/tex]
