Question

an object thrown vertically upward from the surface of a celestial body at a velocity of 27 m/s reaches a height of s = - 0.3t² + 27t meters in t seconds. a. determine the velocity v of the object after t seconds. b. when does the object reach its highest point? c. what is the height of the object at the highest point? d. when does the object strike the ground? e. with what velocity does the object strike the ground? f. on what intervals is the speed increasing?

Explanation:

Step1: Recall velocity - height relationship

The height function is $s(t)= - 0.3t^{2}+27t$. The velocity function $v(t)$ is the derivative of the height function. Using the power - rule $\frac{d}{dt}(at^{n})=nat^{n - 1}$, we have $v(t)=s^\prime(t)=\frac{d}{dt}(-0.3t^{2}+27t)=-0.6t + 27$.

Step2: Find the time at the highest point

At the highest point, the velocity $v(t) = 0$. Set $v(t)=-0.6t + 27 = 0$. Solving for $t$:
\[

$$\begin{align*} -0.6t+27&=0\\ -0.6t&=- 27\\ t&=\frac{-27}{-0.6}=45 \end{align*}$$

\]

Step3: Find the height at the highest point

Substitute $t = 45$ into the height function $s(t)=-0.3t^{2}+27t$.
\[

$$\begin{align*} s(45)&=-0.3\times(45)^{2}+27\times45\\ &=-0.3\times2025 + 1215\\ &=-607.5+1215\\ &=607.5 \end{align*}$$

\]

Step4: Find the time when the object strikes the ground

Set $s(t)=-0.3t^{2}+27t = 0$. Factor out $t$: $t(-0.3t + 27)=0$. We have two solutions: $t = 0$ (corresponds to the initial time) and $-0.3t+27 = 0$. Solving $-0.3t+27 = 0$ gives $t = 90$.

Step5: Find the velocity when the object strikes the ground

Substitute $t = 90$ into the velocity function $v(t)=-0.6t + 27$.
\[

$$\begin{align*} v(90)&=-0.6\times90+27\\ &=-54 + 27\\ &=-27 \end{align*}$$

\]

Step6: Analyze when the speed is increasing

The acceleration $a(t)$ is the derivative of the velocity function. $a(t)=v^\prime(t)=-0.6$. Since the acceleration is negative, the speed $|v(t)|$ is increasing when the velocity and acceleration have the same sign. The velocity $v(t)=-0.6t + 27$ and $a(t)=-0.6$. The velocity is negative when $-0.6t+27<0$, i.e., $t > 45$.

Answer:

a. $v(t)=-0.6t + 27$
b. $t = 45$ seconds
c. $s(45)=607.5$ meters
d. $t = 90$ seconds
e. $v(90)=-27$ m/s
f. $t\in(45,90)$