an object moves along a coordinate line with velocity $v(t)=t(9 - t)$ units per second. its initial position…

an object moves along a coordinate line with velocity $v(t)=t(9 - t)$ units per second. its initial position is 4 units to the left of the origin.\n(a) find the position of the object 15 seconds later.\nanswer: the new position is \nnote: if the object is positioned to the left of the origin at $t = 15$ seconds, then your answer above will be negative.\n(b) find the total distance traveled by the object during those 15 seconds.\nanswer: the object traveled a total distance of

an object moves along a coordinate line with velocity $v(t)=t(9 - t)$ units per second. its initial position is 4 units to the left of the origin.\n(a) find the position of the object 15 seconds later.\nanswer: the new position is \nnote: if the object is positioned to the left of the origin at $t = 15$ seconds, then your answer above will be negative.\n(b) find the total distance traveled by the object during those 15 seconds.\nanswer: the object traveled a total distance of

Answer

Explanation:

Step1: Recall the position - velocity relationship

The position function $s(t)$ is related to the velocity function $v(t)$ by $s(t)-s(0)=\int_{0}^{t}v(u)du$, where $s(0)$ is the initial position. Given $v(t)=t(9 - t)=9t - t^{2}$ and $s(0)=- 4$.

Step2: Calculate the integral for position

First, find $\int_{0}^{15}(9t - t^{2})dt$. Using the power - rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)$, we have $\int_{0}^{15}(9t - t^{2})dt=\left[\frac{9t^{2}}{2}-\frac{t^{3}}{3}\right]_{0}^{15}$.

  • Substitute the upper and lower limits: $\left(\frac{9\times15^{2}}{2}-\frac{15^{3}}{3}\right)-\left(\frac{9\times0^{2}}{2}-\frac{0^{3}}{3}\right)$.
  • Calculate $\frac{9\times15^{2}}{2}=\frac{9\times225}{2}=\frac{2025}{2}$ and $\frac{15^{3}}{3}=\frac{3375}{3}=1125$.
  • Then $\frac{2025}{2}-1125=\frac{2025 - 2250}{2}=-\frac{225}{2}$.
  • Since $s(15)-s(0)=\int_{0}^{15}v(t)dt$, and $s(0)=-4$, then $s(15)=-4-\frac{225}{2}=-\frac{8 + 225}{2}=-\frac{233}{2}$.

Step3: Recall the distance - velocity relationship

The distance $d$ traveled by the object is $d=\int_{0}^{15}|v(t)|dt$. First, find the roots of $v(t)=9t - t^{2}=t(9 - t)$. The roots are $t = 0$ and $t = 9$.

  • For $0\leq t\leq9$, $v(t)\geq0$; for $t>9$, $v(t)<0$.
  • So, $\int_{0}^{15}|v(t)|dt=\int_{0}^{9}(9t - t^{2})dt+\int_{9}^{15}-(9t - t^{2})dt$.
  • Calculate $\int_{0}^{9}(9t - t^{2})dt=\left[\frac{9t^{2}}{2}-\frac{t^{3}}{3}\right]_{0}^{9}=\frac{9\times9^{2}}{2}-\frac{9^{3}}{3}=\frac{9^{3}}{2}-\frac{9^{3}}{3}=\frac{3\times9^{3}-2\times9^{3}}{6}=\frac{9^{3}}{6}=\frac{729}{6}=\frac{243}{2}$.
  • Calculate $\int_{9}^{15}-(9t - t^{2})dt=-\left[\frac{9t^{2}}{2}-\frac{t^{3}}{3}\right]_{9}^{15}=-\left(\left(\frac{9\times15^{2}}{2}-\frac{15^{3}}{3}\right)-\left(\frac{9\times9^{2}}{2}-\frac{9^{3}}{3}\right)\right)$.
    • We already know $\frac{9\times15^{2}}{2}-\frac{15^{3}}{3}=-\frac{225}{2}$ and $\frac{9\times9^{2}}{2}-\frac{9^{3}}{3}=\frac{243}{2}$.
    • So, $-\left(-\frac{225}{2}-\frac{243}{2}\right)=\frac{225 + 243}{2}=\frac{468}{2}=234$.
  • Then $d=\frac{243}{2}+234=\frac{243 + 468}{2}=\frac{711}{2}$.

Answer:

(a) $-\frac{233}{2}$ (b) $\frac{711}{2}$