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

Explanation:

Step1: Find velocity function

The velocity $v(t)$ is the derivative of the position function $s(t)=t^{3}-3t^{2}+3t$. Using the power - rule $\frac{d}{dt}(t^{n}) = nt^{n - 1}$, we have $v(t)=s^\prime(t)=3t^{2}-6t + 3$.

Step2: Find acceleration function

The acceleration $a(t)$ is the derivative of the velocity function. So $a(t)=v^\prime(t)=6t-6$.

Step3: Calculate displacement

Displacement $\Delta s=s(4)-s(0)$. [ \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.

Step4: Calculate average velocity

Average velocity $v_{avg}=\frac{s(4)-s(0)}{4 - 0}=\frac{28}{4}=7$ m/s.

Step5: Calculate speed at endpoints

Speed is the absolute value of velocity. At $t = 0$, $v(0)=3\times0^{2}-6\times0 + 3=3$ m/s. At $t = 4$, $v(4)=3\times4^{2}-6\times4 + 3=48-24 + 3=27$ m/s.

Step6: Calculate acceleration at endpoints

At $t = 0$, $a(0)=6\times0-6=-6$ m/s². At $t = 4$, $a(4)=6\times4-6=18$ m/s².

Step7: Find when the body changes direction

The body changes direction when $v(t)$ changes sign. Set $v(t)=3t^{2}-6t + 3 = 0$. Factor out a 3: $3(t^{2}-2t + 1)=0$, or $3(t - 1)^{2}=0$. The solution is $t = 1$. 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 $t = 2$, $v(2)=3\times2^{2}-6\times2 + 3=12-12 + 3=3>0$. The body changes direction at $t = 1$ s.

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. The body changes direction at $t = 1$ s