Answer
(a) $v(t)=0.0015~t^3-0.1155~t^2+24.9817~t-21.2687$
We can see a sketch of this graph in the interval $[0, 125]$
(b) $a(t) = 0.0045~t^2-0.231~t+24.9817$
During the first 125 seconds:
The minimum acceleration is $22.0~ft/s^2$
The maximum acceleration is $66.4~ft/s^2$
Work Step by Step
(a) Using the cubic regression function on a calculator, we can use the data in the table to find the following cubic polynomial:
$v(t)=0.0015~t^3-0.1155~t^2+24.9817~t-21.2687$
We can see a sketch of this graph in the interval $[0, 125]$
(b) We can find an equation for $a(t)$:
$a(t) = v'(t) = 0.0045~t^2-0.231~t+24.9817$
We can find the time $t$ when $a'(t) = 0$:
$a'(t) = 0.009~t-0.231 = 0$
$t = \frac{0.231}{0.009}$
$t = 25.67~s$
When $t = 0$:
$a(0) = 0.0045~(0)^2-0.231~(0)+24.9817 = 25.0~ft/s^2$
When $t = 25.67$:
$a(25.67) = 0.0045~(25.67)^2-0.231~(25.67)+24.9817 = 22.0~ft/s^2$
When $t = 125$:
$a(125) = 0.0045~(125)^2-0.231~(125)+24.9817 = 66.4~ft/s^2$
During the first 125 seconds:
The minimum acceleration is $22.0~ft/s^2$
The maximum acceleration is $66.4~ft/s^2$
