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

Velocity $v(t)=s^\prime(t)$, acceleration $a(t)=v^\prime(t)=s^{\prime\prime}(t)$. Speed is $|v(t)|$. Displacement is $s(b)-s(a)$ and average - velocity is $\frac{s(b)-s(a)}{b - a}$ for the interval $[a,b]$. The body changes direction when $v(t)$ changes sign. First, find the derivative of $s(t)=t^{3}-3t^{2}+3t$. Using the power rule $(x^n)^\prime=nx^{n - 1}$, we have $v(t)=s^\prime(t)=3t^{2}-6t + 3$. The second - derivative $a(t)=v^\prime(t)=s^{\prime\prime}(t)=6t-6$.

Step2: Calculate displacement and average velocity for $[0,4]$

Displacement: $s(4)=4^{3}-3\times4^{2}+3\times4=64 - 48+12 = 28$. $s(0)=0^{3}-3\times0^{2}+3\times0 = 0$. Displacement $=s(4)-s(0)=28$ meters. Average velocity $=\frac{s(4)-s(0)}{4 - 0}=\frac{28}{4}=7$ m/s.

Step3: Calculate speed and acceleration at endpoints

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

Step4: Find when the body changes direction

Set $v(t)=3t^{2}-6t + 3 = 0$. Factor out 3: $3(t^{2}-2t + 1)=0$. Since $t^{2}-2t + 1=(t - 1)^{2}$, then $(t - 1)^{2}=0$, so $t = 1$. We can check the sign of $v(t)$ on intervals. For $t\in[0,1)$, let $t = 0.5$, $v(0.5)=3\times(0.5)^{2}-6\times0.5 + 3=3\times0.25-3 + 3=0.75>0$. For $t\in(1,4]$, let $t = 2$, $v(2)=3\times2^{2}-6\times2 + 3=12-12 + 3=3>0$. But $v(t)$ is zero at $t = 1$, so the body changes direction at $t = 1$ s.

Answer:

a. Displacement: 28 meters, 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. A. $t = 1$ sec