Answer
(a) $x(t) = (0.250~m/s^3)~t^3 - (0.0100~m/s^4)~t^4$
$v_x(t) = (0.750~m/s^3)~t^2 - (0.0400~m/s^4)~t^3$
(b) The maximum velocity is 39.1 m/s
Work Step by Step
$a_x(t) = At - Bt^2$
$a_x(t) = (1.50~m/s^3)~t - (0.120~m/s^4)~t^2$
Note that $v_{0x} = 0$ and $x_0 = 0 $ because initially the motorcycle is at rest at the origin.
(a) $v_x(t) = v_{0x}+ \int_{0}^{t}a_x(t)~dt$
$v_x(t) = 0+ \int_{0}^{t} (1.50~m/s^3)~t - (0.120~m/s^4)~t^2~dt$
$v_x(t) = (0.750~m/s^3)~t^2 - (0.0400~m/s^4)~t^3$
We can use $v_x(t)$ to find $x(t)$.
$x(t) = x_0+ \int_{0}^{t}v_x(t)~dt$
$x(t) = 0+ \int_{0}^{t}(0.750~m/s^3)~t^2 - (0.0400~m/s^4)~t^3~dt$
$x(t) = (0.250~m/s^3)~t^3 - (0.0100~m/s^4)~t^4$
(b) The maximum velocity occurs at the moment when $a_x(t) = 0$. It is after this moment that the velocity starts to decrease.
We can find time $t$ when $a_x(t) = 0$.
$a_x(t) = (1.50~m/s^3)~t - (0.120~m/s^4)~t^2 = 0$
$(t)((1.50~m/s^3) - (0.120~m/s^4)~t) = 0$
$t=0$ or $(1.50~m/s^3) - (0.120~m/s^4)~t = 0$
$(1.50~m/s^3) = (0.120~m/s^4)~t$
$t = \frac{1.50~m/s^3}{0.120~m/s^4} = 12.5~s$
At t = 12.5 s,
$v = (0.750~m/s^3)(12.5~s)^2 - (0.0400~m/s^4)(12.5~s)^3$
$v = 39.1~m/s$
The maximum velocity is 39.1 m/s
