Answer
(a) $C = \frac{5}{8}~m/s^3$
(b) The object travels a distance of 106.7 meters during the 8.00 seconds.
Work Step by Step
$a_x(t) = -Ct$
We can use $a_x(t)$ to find $v_x(t)$.
$v_x(t) = v_{0x}+ \int_{o}^{t}a_x(t)~dt$
$v_x(t)=v_{0x}+ \int_{o}^{t}-Ct~dt$
$v_x(t)= (20~m/s)+(\frac{-Ct^2}{2})$
At t = 8.00 s,
$v= (20~m/s)+(\frac{-C}{2})~t^2 = 0$
$C = \frac{(2)(20~m/s)}{t^2} = \frac{(2)(20~m/s)}{(8.00~s)^2}$
$C = \frac{5}{8}~m/s^3$
(b) We can use $v_x(t)$ to find $x(t)$.
$x(t) = x_0+ \int_{0}^{t}v_x(t)~dt$
$x(t) = x_0+ \int_{0}^{t}(20~m/s)+(\frac{-5}{16}~m/s^3)~t^2~dt$
$x(t) = x_0+ (20~m/s)~t+(\frac{-5}{48}~m/s^3)~t^3$
When t = 0,
$x = x_0$
When t = 8.00 s,
$x = x_0+ (20~m/s)(8.00~s)+(\frac{-5}{48}~m/s^3)(8.00~s)^3$
$x = x_0 + 106.7~m$
The object travels a distance of 106.7 meters during the 8.00 seconds.
