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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?

Explanation:

Step1: Recall velocity - displacement formula

The height - time function is given by $s = 40t-0.8t^{2}$. The velocity function $v$ is the derivative of the position function $s$ with respect to time $t$. Using the power rule $\frac{d}{dt}(t^{n})=nt^{n - 1}$, we have $v=\frac{ds}{dt}=40 - 1.6t$.

Step2: Find time to reach maximum height

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

Step3: Find maximum height

Substitute $t = 25$ s into the position function $s = 40t-0.8t^{2}$. So $s=40\times25-0.8\times25^{2}=1000 - 0.8\times625=1000 - 500 = 500$ m.

Step4: Find time to reach half - maximum height

Set $s = 250$ (half of 500). So $250=40t-0.8t^{2}$, which can be rewritten as $0.8t^{2}-40t + 250 = 0$. 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 $c = 625$. So $t=\frac{100\pm\sqrt{(-100)^{2}-4\times2\times625}}{2\times2}=\frac{100\pm\sqrt{10000 - 5000}}{4}=\frac{100\pm\sqrt{5000}}{4}=\frac{100\pm50\sqrt{2}}{4}=\frac{50\pm25\sqrt{2}}{2}$. The two solutions are $t_1=\frac{50 + 25\sqrt{2}}{2}\approx\frac{50+25\times1.414}{2}=\frac{50 + 35.35}{2}=42.675$ s and $t_2=\frac{50 - 25\sqrt{2}}{2}\approx\frac{50-35.35}{2}=7.325$ s. The smaller value is the time to reach the first - half of the maximum height on the way up.

Step5: Find total time aloft

Set $s = 0$. So $0=40t-0.8t^{2}=t(40 - 0.8t)$. This gives two solutions: $t = 0$ (corresponds to the initial time) and $40-0.8t=0$, which gives $t = 50$ s.

Answer:

a. $v = 40 - 1.6t$ m/s
b. $t = 25$ s
c. $s = 500$ m
d. $t=\frac{50 - 25\sqrt{2}}{2}\approx7.325$ s
e. $t = 50$ s