Answer
The cyclist traveled $~~82.8~m$
Work Step by Step
$a = (6.1-1.2t)~m/s^2$
$v = (6.1t-0.6t^2+2.7)~m/s$
$x = (3.05t^2-0.2t^3+2.7t+7.3)~m$
On the interval $0 \leq t \leq 6.0$, note that $v \gt 0$, so the cyclist continues moving in the positive x-direction on this interval.
When $t = 0$, then $x = 7.3~m$
When $t = 6.0$:
$x = (3.05t^2-0.2t^3+2.7t+7.3)~m$
$x = [3.05(6.0)^2-0.2(6.0)^3+2.7(6.0)+7.3]~m$
$x = 90.1~m$
We can find the distance the cyclist traveled:
$90.1~m - 7.3~m = 82.8~m$
The cyclist traveled $~~82.8~m$
