Question

a rock thrown vertically upward from the surface of the moon at a velocity of 8 m/sec reaches a height of s = 8t - 0.8t² meters in t sec. a. find the rocks velocity and acceleration at time t. b. how long does it take the rock to reach its highest point? c. how high does the rock go? d. how long does it take the rock to reach half its maximum height? e. how long is the rock aloft?

Explanation:

Step1: Find velocity and acceleration

Velocity $v(t)$ is the derivative of position $s(t)$. Given $s(t)=8t - 0.8t^{2}$, by the power - rule $\frac{d}{dt}(at^{n})=nat^{n - 1}$, we have $v(t)=s^\prime(t)=8 - 1.6t$. Acceleration $a(t)$ is the derivative of velocity. So $a(t)=v^\prime(t)=- 1.6$.

Step2: Find time to reach highest point

At the highest point, $v(t)=0$. Set $8 - 1.6t = 0$. Solving for $t$ gives $1.6t=8$, so $t = 5$ s.

Step3: Find maximum height

Substitute $t = 5$ into $s(t)$. $s(5)=8\times5-0.8\times5^{2}=40 - 20=20$ m.

Step4: Find time to reach half - maximum height

Half of the[SSE Completed, Client Connection Error][SSE Completed, Client Connection Error][LLM SSE On Failure]

Answer:

Step1: Find velocity and acceleration

Velocity $v(t)$ is the derivative of position $s(t)$. Given $s(t)=8t - 0.8t^{2}$, by the power - rule $\frac{d}{dt}(at^{n})=nat^{n - 1}$, we have $v(t)=s^\prime(t)=8 - 1.6t$. Acceleration $a(t)$ is the derivative of velocity. So $a(t)=v^\prime(t)=- 1.6$.

Step2: Find time to reach highest point

At the highest point, $v(t)=0$. Set $8 - 1.6t = 0$. Solving for $t$ gives $1.6t=8$, so $t = 5$ s.

Step3: Find maximum height

Substitute $t = 5$ into $s(t)$. $s(5)=8\times5-0.8\times5^{2}=40 - 20=20$ m.

Step4: Find time to reach half - maximum height

Half of the[SSE Completed, Client Connection Error][SSE Completed, Client Connection Error][LLM SSE On Failure]