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

a) The velocity is $0$ when $t=1$ and $t=3$. $a(1)=-6m/s^2$ and $a(3)=6m/s^2$. b) The acceleration is $0$ when $t=2$. The speed of the body at $t=2$ is $3m/s$. c) The total distance traveled by the body is $6m$.
$$s(t)=t^3-6t^2+9t$$ First, we need to find $v(t)$ and $a(t)$. $$v(t)=\frac{ds}{dt}=\frac{d}{dt}(t^3-6t^2+9t)=3t^2-12t+9=3(t^2-4t+3)$$ $$v(t)=3(t-1)(t-3)$$ $$a(t)=\frac{dv}{dt}=\frac{d}{dt}(3t^2-12t+9)=6t-12$$ a) The velocity is $0$ when $v(t)=0$, which are when either $t=1$ or $t=3$. - For $t=1$: $a(1)=6\times1-12=-6m/s^2$ - For $t=3$: $a(3)=6\times3-12=18-12=6m/s^2$ b) The acceleration is $0$ when $a(t)=0$, which is when $t=2$. - For $t=2$: $v(2)=3(2-1)(2-3)=3\times1\times(-1)=-3m/s$ So the speed of the body at $t=2$ is $|v(2)|=3m/s$. c) To find the total distance traveled by the body means to take into account the sum of both the distance traveled by the body forward and backward. First, we need to find out at which interval in $[0,2]$ the body travels forward and backward. From part a), we figure that, on $[0,2]$, at $t=1$, $v(t)=0$. - On $[0,1)$, $(t-1)\lt0$ while $(t-3)\lt0$, so $v(t)\gt0$ and the body travels forward. - On $(1,2]$, $(t-1)\gt0$ while $(t-3)\lt0$, so $v(t)\lt0$ and the body travels backward. Now, to find the total distance traveled by the body from $t=0$ to $t=2$, we take the sum of the distance the body travels forward and backward. $$s(1)-s(0)+|s(2)-s(1)|=(1^3-6\times1^2+9\times1)-(0^3-6\times0^2+9\times0)+|(2^3-6\times2^2+9\times2)-(1^3-6\times1^2+9\times1)|$$ $$s(1)-s(0)+|s(2)-s(1)|=4-0+|2-4|=4+|-2|=6m$$