Question

a rock thrown vertically upward from the surface of the moon at a velocity of 20 m/sec reaches a height of s = 20t - 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?
find the rocks acceleration at time t
a = - 1.6 m/s²
(simplify your answer. use integers or decimals for any numbers in the expression.)
b. how long does it take the rock to reach its highest point?
12.5 sec
(simplify your answer.)
c. how high does the rock go?
125 m
(simplify your answer.)
d. how long does it take the rock to reach half its maximum height?
sec
(simplify your answer. round to two decimal places as needed. use a comma to separate answers as needed.)

Explanation:

Step1: Recall velocity - height relationship

The height function is $s(t)=20t - 0.8t^{2}$. 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 $v(t)=s^\prime(t)=\frac{d}{dt}(20t - 0.8t^{2})=20-1.6t$.

Step2: Recall acceleration - velocity relationship

The acceleration function $a(t)$ is the derivative of the velocity function. So $a(t)=v^\prime(t)=\frac{d}{dt}(20 - 1.6t)=-1.6$ m/s².

Step3: Find time to reach highest point

At the highest point, the velocity $v(t) = 0$. Set $v(t)=20 - 1.6t=0$. Solving for $t$ gives $1.6t = 20$, so $t=\frac{20}{1.6}=12.5$ s.

Step4: Find maximum height

Substitute $t = 12.5$ s into the height function $s(t)$. $s(12.5)=20\times12.5-0.8\times(12.5)^{2}=250 - 0.8\times156.25=250 - 125 = 125$ m.

Step5: Find time to reach half - maximum height

The maximum height is 125 m, so half - maximum height is $s = 62.5$ m. Set $s(t)=20t-0.8t^{2}=62.5$. Rearrange to get $0.8t^{2}-20t + 62.5 = 0$. Multiply through by 10 to clear the decimal: $8t^{2}-200t + 625 = 0$. Using the quadratic formula $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for the quadratic equation $ax^{2}+bx + c = 0$, here $a = 8$, $b=-200$, $c = 625$. Then $t=\frac{200\pm\sqrt{(-200)^{2}-4\times8\times625}}{2\times8}=\frac{200\pm\sqrt{40000 - 20000}}{16}=\frac{200\pm\sqrt{20000}}{16}=\frac{200\pm100\sqrt{2}}{16}=\frac{25\pm12.5\sqrt{2}}{2}\approx3.47,21.53$ s.

Step6: Find time the rock is aloft

The rock is aloft when it returns to the surface, i.e., $s(t)=0$. Set $20t-0.8t^{2}=0$. Factor out $t$: $t(20 - 0.8t)=0$. We get $t = 0$ (corresponds to the time of throwing) and $20-0.8t=0$, which gives $t = 25$ s.

Answer:

a. $v(t)=20 - 1.6t$ m/s, $a(t)=-1.6$ m/s²
b. $12.5$ s
c. $125$ m
d. $3.47$ s, $21.53$ s
e. $25$ s