Question

a rock thrown vertically upward from the surface of the moon at a velocity of 16 m/sec reaches a height of s = 16t - 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 = m/s

Explanation:

Step1: Recall velocity - derivative of position

Velocity $v$ is the derivative of the position function $s(t)$. Given $s(t)=16t - 0.8t^{2}$, using the power - rule $\frac{d}{dt}(at^{n})=nat^{n - 1}$, we have $v(t)=\frac{d}{dt}(16t-0.8t^{2})$.
$v(t)=\frac{d}{dt}(16t)-\frac{d}{dt}(0.8t^{2})$.
$v(t)=16 - 1.6t$.

Step2: Recall acceleration - derivative of velocity

Acceleration $a$ is the derivative of the velocity function. So $a(t)=\frac{d}{dt}(v(t))$. Since $v(t)=16 - 1.6t$, then $a(t)=\frac{d}{dt}(16)-\frac{d}{dt}(1.6t)$.
$a(t)=0 - 1.6=-1.6$.

Step3: Find time to reach highest point

At the highest point, velocity $v = 0$. Set $v(t)=0$.
$16 - 1.6t = 0$.
$1.6t = 16$, so $t=\frac{16}{1.6}=10$ s.

Step4: Find maximum height

Substitute $t = 10$ into the position function $s(t)$.
$s(10)=16\times10-0.8\times10^{2}=160 - 80 = 80$ m.

Step5: Find time to reach half - maximum height

Half - maximum height is $s = 40$ m. Set $s(t)=40$, so $16t-0.8t^{2}=40$.
Rearrange to get $0.8t^{2}-16t + 40 = 0$. Multiply through by 10 to clear the decimal: $8t^{2}-160t + 400 = 0$. Divide by 8: $t^{2}-20t + 50 = 0$.
Using the quadratic formula $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ with $a = 1$, $b=-20$, and $c = 50$.
$t=\frac{20\pm\sqrt{(-20)^{2}-4\times1\times50}}{2\times1}=\frac{20\pm\sqrt{400 - 200}}{2}=\frac{20\pm\sqrt{200}}{2}=\frac{20\pm10\sqrt{2}}{2}=10\pm5\sqrt{2}$.

Step6: Find time the rock is aloft

The rock is aloft when $s(t)=0$. Set $s(t)=16t-0.8t^{2}=0$.
Factor out $t$: $t(16 - 0.8t)=0$. So $t = 0$ (initial time) or $16-0.8t = 0$.
$0.8t = 16$, $t = 20$ s.

Answer:

a. $v = 16 - 1.6t$ m/s, $a=-1.6$ m/s²
b. $t = 10$ s
c. $s = 80$ m
d. $t = 10\pm5\sqrt{2}$ s
e. $t = 20$ s