QUESTION IMAGE
Question
a rock thrown vertically upward from the surface of the moon at a velocity of 8 m/sec reaches a height of $s = 8t - 0.8t^2$ 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.
$v = \square$ m/s
Step1: Differentiate height for velocity
Velocity $v(t)$ is the derivative of height $s(t)=8t-0.8t^2$. Using power rule: $\frac{d}{dt}(t^n)=nt^{n-1}$.
$v(t)=\frac{d}{dt}(8t-0.8t^2)=8 - 1.6t$
Step2: Differentiate velocity for acceleration
Acceleration $a(t)$ is the derivative of velocity $v(t)=8-1.6t$.
$a(t)=\frac{d}{dt}(8-1.6t)=-1.6$
Step3: Find time at highest point
At highest point, $v(t)=0$. Solve $8-1.6t=0$.
$1.6t=8 \implies t=\frac{8}{1.6}=5$
Step4: Calculate maximum height
Substitute $t=5$ into $s(t)$.
$s(5)=8(5)-0.8(5)^2=40-0.8(25)=40-20=20$
Step5: Find time for half max height
Half max height is $\frac{20}{2}=10$. Solve $8t-0.8t^2=10$.
Rearrange: $0.8t^2-8t+10=0$. Multiply by 10: $8t^2-80t+100=0$. Divide by 4: $2t^2-20t+25=0$. Use quadratic formula $t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ where $a=2, b=-20, c=25$.
$t=\frac{20\pm\sqrt{400-200}}{4}=\frac{20\pm\sqrt{200}}{4}=\frac{20\pm10\sqrt{2}}{4}=\frac{10\pm5\sqrt{2}}{2}$
Simplify: $t=5-\frac{5\sqrt{2}}{2}\approx1.46$ (ascending) and $t=5+\frac{5\sqrt{2}}{2}\approx8.54$ (descending)
Step6: Find time aloft
Rock is aloft when $s(t)=0$. Solve $8t-0.8t^2=0$.
$t(8-0.8t)=0$. Solutions: $t=0$ (launch) and $8-0.8t=0 \implies t=\frac{8}{0.8}=10$
Step1: Differentiate height function
Velocity is derivative of $s(t)=8t-0.8t^2$.
$v(t)=\frac{d}{dt}(8t-0.8t^2)=8-1.6t$
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a. Velocity: $v = 8 - 1.6t$ m/s; Acceleration: $a = -1.6$ m/s²
b. $5$ sec
c. $20$ meters
d. $t=5-\frac{5\sqrt{2}}{2}$ sec (ascending) and $t=5+\frac{5\sqrt{2}}{2}$ sec (descending)
e. $10$ sec
For the specific fill-in part (a, velocity only):