the function s = t³ - 3t² + 3t, 0 ≤ t ≤ 4, gives the position of a body moving on a coordinate line, with s…

Explanation:

Step1: Recall relevant formulas

The velocity function $v(t)$ is the derivative of the position function $s(t)$, and the acceleration function $a(t)$ is the derivative of the velocity function. The body changes direction when $v(t)$ changes sign. Given $s(t)=t^{3}-3t^{2}+3t$, $v(t)=s^\prime(t)=3t^{2}-6t + 3$ and $a(t)=v^\prime(t)=6t-6$.

Step2: Calculate displacement and average - velocity

The displacement over the interval $[0,4]$ is $\Delta s=s(4)-s(0)$. [ \begin{align*} s(4)&=4^{3}-3\times4^{2}+3\times4=64 - 48+12=28\ s(0)&=0^{3}-3\times0^{2}+3\times0 = 0 \end{align*} ] So, $\Delta s=28 - 0=28$ m. The average - velocity $v_{avg}=\frac{\Delta s}{\Delta t}=\frac{s(4)-s(0)}{4 - 0}=\frac{28}{4}=7$ m/s.

Step3: Calculate speed and acceleration at endpoints

For speed at $t = 0$ and $t = 4$:

• $v(t)=3t^{2}-6t + 3$, so $v(0)=3\times0^{2}-6\times0 + 3=3$ m/s, $v(4)=3\times4^{2}-6\times4 + 3=48-24 + 3=27$ m/s.
• $a(t)=6t - 6$, so $a(0)=6\times0-6=-6$ m/s², $a(4)=6\times4-6=18$ m/s².

Step4: Find when the body changes direction

Set $v(t)=3t^{2}-6t + 3 = 0$. Factor out 3: $3(t^{2}-2t + 1)=0$, which is $3(t - 1)^{2}=0$. The root is $t = 1$. But $v(t)=3(t - 1)^{2}\geq0$ for all $t$. So the body does not change direction in the interval $[0,4]$.

Answer:

a. Displacement: 28 m, Average - velocity: 7 m/s b. Speed at $t = 0$: 3 m/s, Speed at $t = 4$: 27 m/s, Acceleration at $t = 0$: - 6 m/s², Acceleration at $t = 4$: 18 m/s² c. B. The body does not change direction during the interval