## University Calculus: Early Transcendentals (3rd Edition)

a) The body's displacement is $2.25m$ and the body's average speed is $0.75m/s$. b) At $t=0$: speed $=0m/s$, acceleration $=2m/s^2$ At $t=3$: speed $=6m/s$, acceleration $=11m/s^2$ c) The body changes directions at 2 points in the interval: $t=1s$ and $t=2s$.
$$s(t)=\frac{t^4}{4}-t^3+t^2$$ a) The body's displacement, $\Delta s$, over the interval $[0,3]$ is $$\Delta s=s(3)-s(0)=(\frac{3^4}{4}-3^3+3^2)-(\frac{0^4}{4}-0^3+0^2)$$ $$\Delta s=\frac{81}{4}-27+9-0=2.25m$$ The average velocity, $v_a$, of the body over the interval $[0,3]$ is $$v_a=\frac{\Delta s}{3-0}=\frac{2.25}{3}=0.75m/s$$ b) We first look for the velocity of the body at time $t$, $v(t)$, and the acceleration of the body at time $t$, $a(t)$: $$v(t)=\frac{ds}{dt}=\frac{d}{dt}\Big(\frac{t^4}{4}-t^3+t^2\Big)=t^3-3t^2+2t$$ $$a(t)=\frac{dv}{dt}=\frac{d}{dt}(t^3-3t^2+2t)=3t^2-6t+2$$ We now examine the speed and acceleration at 2 endpoints of $[0,3]$: - At $t=0$: $$v(0)=0^3-3\times0^2+2\times0=0m/s$$ So the speed of the body is $|v(0)|=0m/s$ $$a(0)=-3\times0^2-6\times0+2=2m/s^2$$ - At $t=3$: $$v(3)=3^3-3\times3^2+2\times3=6m/s$$ So the speed of the body is $|v(3)|=6m/s$ $$a(3)=3\times3^2-6\times3+2=27-18+2=11m/s^2$$ c) We know that if - $v(t)\gt0$: the body moves forward. - $v(t)\lt0$: the body moves backward. So the body will change direction when $v(t)$ changes sign. We will look for points in $[0,3]$ where $v(t)=0$, then examine the signs of $v(t)$ before and after that point to determine whether $v(t)$ changes sign there. We have $$v(t)=t^3-3t^2+2t=t(t^2-3t+2)=t(t-1)(t-2)$$ So $v(t)=0$ if $t=0s$, $t=1s$ or $t=2s$. We do not examine $t=0s$ because that is the starting endpoint of the interval, so there is no "changing direction" for the body there. - For $t=1s$: On $[0,1)$, $t\gt0$, $(t-1)\lt0$, $(t-2)\lt0$; therefore, $v(t)\gt0$. On $(1,2)$, $t\gt0$, $(t-1)\gt0$, $(t-2)\lt0$; therefore, $v(t)\lt0$. So $v(t)$ changes sign at $t=1$. - For $t=2s$: On $(1,2)$, we have shown that $v(t)\lt0$. On $(2,3]$, $t\gt0$, $(t-1)\gt0$, $(t-2)\gt0$; therefore, $v(t)\gt0$. So $v(t)$ changes sign at $t=2$. That means the body changes directions at 2 points in the interval: $t=1s$ and $t=2s$.