Answer:
(a) Velocity = [tex]3t^2-16t+23[/tex]
(b) Velocity after 1 second is 10 ft/s
Explanation:
We know that if we differentiate distance with respect to time, we get velocity.
Thus we can use the same concept with an equation for distance, and solve the question as follows:
(a) We simply differentiate the following equation: [tex]f(t) = t^3 - 8t^2 +23t[/tex]
After differentiating, this becomes: [tex]\frac{f(t)}{dt} = 3t^2-16t+23[/tex]
This is also the new equation for velocity.
So, we have: Velocity = [tex]3t^2-16t+23[/tex]
(b) At time = 1 s, we have the velocity (using our velocity equation from part 1):
[tex]v(1) = 3(1^2)-16(1)+23[/tex]
[tex]v(1) = 10 ft/s[/tex]
So velocity after 1 second is 10 ft/s
