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a rock thrown vertically upward from the surface of the moon at a veloc…

Question

a rock thrown vertically upward from the surface of the moon at a velocity of 40 m/sec reaches a height of s = 40t - 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? a. find the rocks velocity at time t. v = 40 - 1.6t m/s find the rocks acceleration at time t. a = 25 m/s² (simplify your answer. use integers or decimals for any numbers in the expression.)

Explanation:

Step1: Recall velocity - derivative of position

Velocity $v$ is the derivative of the position function $s(t)$. Given $s = 40t-0.8t^{2}$, using the power - rule $\frac{d}{dt}(at^{n})=nat^{n - 1}$, we have $v=\frac{ds}{dt}=\frac{d}{dt}(40t-0.8t^{2})=40 - 1.6t$ m/s.

Step2: Recall acceleration - derivative of velocity

Acceleration $a$ is the derivative of the velocity function $v(t)$. Since $v = 40 - 1.6t$, then $a=\frac{dv}{dt}=\frac{d}{dt}(40 - 1.6t)=- 1.6$ m/s². The value of $a = 25$ in the given problem is incorrect.

Answer:

a. Velocity: $v = 40 - 1.6t$ m/s, Acceleration: $a=-1.6$ m/s²
b. At the highest - point, $v = 0$. So, $40-1.6t = 0$. Solving for $t$ gives $t=\frac{40}{1.6}=25$ s.
c. Substitute $t = 25$ s into the position function $s(t)$. $s(25)=40\times25-0.8\times25^{2}=1000 - 0.8\times625=1000 - 500 = 500$ m.
d. First, the maximum height is $s_{max}=500$ m. Half of the maximum height is $s = 250$ m. So, $250=40t-0.8t^{2}$. Rearranging gives $0.8t^{2}-40t + 250 = 0$. Using the quadratic formula $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for the quadratic equation $ax^{2}+bx + c = 0$ (here $a = 0.8$, $b=-40$, $c = 250$), $t=\frac{40\pm\sqrt{(-40)^{2}-4\times0.8\times250}}{2\times0.8}=\frac{40\pm\sqrt{1600 - 800}}{1.6}=\frac{40\pm\sqrt{800}}{1.6}=\frac{40\pm20\sqrt{2}}{1.6}=\frac{40\pm28.28}{1.6}$. We get $t_1=\frac{40 + 28.28}{1.6}\approx42.7$ s and $t_2=\frac{40 - 28.28}{1.6}\approx7.3$ s.
e. The rock is aloft when $s(t)=0$. So, $40t-0.8t^{2}=0$. Factoring out $t$ gives $t(40 - 0.8t)=0$. 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