## University Physics with Modern Physics (14th Edition)

(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
$a_x(t) = At - Bt^2$ $a_x(t) = (1.50~m/s^3)~t - (0.120~m/s^4)~t^2$ 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