Question

a rock thrown vertically upward from the surface of the mo 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 e. how long is the rock aloft? a. find the rocks velocity at time t. v = 40 - 1.6t m/s find the rocks acceleration at time t. a = 14.4 m/s² (simplify your answer. use integers or decimals for any num)

Explanation:

Step1: Recall velocity - time and acceleration - time relationships

The velocity function $v(t)$ is given as $v = 40 - 1.6t$ m/s. The acceleration $a(t)$ is the derivative of the velocity function. The derivative of a linear function $y=mx + b$ is $y'=m$. For $v(t)=40 - 1.6t$, the derivative with respect to $t$ is $a(t)=\frac{d}{dt}(40 - 1.6t)$.

Step2: Calculate the derivative of the velocity function

Using the power - rule of differentiation $\frac{d}{dt}(C)=0$ (where $C$ is a constant) and $\frac{d}{dt}(kt)=k$ (where $k$ is a constant), we have $\frac{d}{dt}(40 - 1.6t)=\frac{d}{dt}(40)-\frac{d}{dt}(1.6t)=0 - 1.6=- 1.6$ m/s². It seems there is an error in the provided acceleration value of $14.4$ m/s². The correct acceleration for a vertically - thrown object near the surface of the moon (assuming a constant acceleration due to gravity) should be the negative of the gravitational acceleration on the moon. The gravitational acceleration on the moon is approximately $1.6$ m/s², and since the object is moving against gravity, $a=-1.6$ m/s².

Step3: Find the time to reach the highest point

At the highest point, the velocity $v = 0$. Set $v(t)=40 - 1.6t = 0$. Solving for $t$ gives $1.6t=40$, so $t=\frac{40}{1.6}=25$ s.

Step4: Find the maximum height

We know that the position function $s(t)$ can be found by integrating the velocity function. $v(t)=40 - 1.6t$, so $s(t)=\int(40 - 1.6t)dt=40t-0.8t^{2}+C$. Assuming the initial position $s(0) = 0$ (thrown from the surface, so $C = 0$). Substitute $t = 25$ s (the time to reach the highest point) into $s(t)$: $s(25)=40\times25-0.8\times25^{2}=1000 - 0.8\times625=1000 - 500 = 500$ m.

Step5: Find the time to reach half - maximum height

The maximum height is $s_{max}=500$ m. We want to find $t$ when $s(t)=250$ m. So, $40t-0.8t^{2}=250$. Rearranging gives $0.8t^{2}-40t + 250 = 0$. Multiply through by 10 to get $8t^{2}-400t + 2500 = 0$, or $2t^{2}-100t + 625 = 0$. Using the quadratic formula $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for the quadratic equation $ax^{2}+bx + c = 0$, here $a = 2$, $b=-100$, and $c = 625$. $t=\frac{100\pm\sqrt{(-100)^{2}-4\times2\times625}}{2\times2}=\frac{100\pm\sqrt{10000 - 5000}}{4}=\frac{100\pm\sqrt{5000}}{4}=\frac{100\pm50\sqrt{2}}{4}=\frac{50(2\pm\sqrt{2})}{4}=\frac{25(2\pm\sqrt{2})}{2}$. We take the smaller value of $t$ for the upward motion, $t=\frac{25(2 - \sqrt{2})}{2}\approx\frac{25(2 - 1.414)}{2}=\frac{25\times0.586}{2}\approx7.325$ s.

Step6: Find the time the rock is aloft

The rock returns to the starting position when $s(t)=0$. So, $40t-0.8t^{2}=0$. Factor out $t$: $t(40 - 0.8t)=0$. We have two solutions: $t = 0$ (corresponds to the initial time) and $40-0.8t=0$, which gives $t = 50$ s.

Answer:

a. Velocity $v = 40 - 1.6t$ m/s, Acceleration $a=-1.6$ m/s²
b. $t = 25$ s
c. $h = 500$ m
d. $t\approx7.325$ s
e. $t = 50$ s