Question

1. if a rock is thrown vertically upward on mars with an initial velocity of 20 ft/sec, its height in feet t seconds later is given by s = 20t - 6t².

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 at its highest point?
d) how long does it take for the rock to reach half of its maximum height?
e) how long is the rock aloft?

Explanation:

Step1: Find velocity by differentiating height function

The height function is $s(t)=20t - 6t^{2}$. The velocity function $v(t)$ is the derivative of the height function. Using the power - rule $\frac{d}{dt}(t^n)=nt^{n - 1}$, we have $v(t)=\frac{d}{dt}(20t-6t^{2})=20 - 12t$. The acceleration function $a(t)$ is the derivative of the velocity function. So $a(t)=\frac{d}{dt}(20 - 12t)=- 12$.

Step2: Find time to reach highest point

At the highest point, the velocity is 0. Set $v(t)=0$. So $20 - 12t = 0$. Solving for $t$ gives $12t=20$, and $t=\frac{20}{12}=\frac{5}{3}$ seconds.

Step3: Find maximum height

Substitute the time $t = \frac{5}{3}$ into the height function $s(t)$. $s(\frac{5}{3})=20\times\frac{5}{3}-6\times(\frac{5}{3})^{2}=\frac{100}{3}-6\times\frac{25}{9}=\frac{100}{3}-\frac{50}{3}=\frac{50}{3}$ feet.

Step4: Find time to reach half - maximum height

Half of the maximum height is $\frac{1}{2}\times\frac{50}{3}=\frac{25}{3}$ feet. Set $s(t)=\frac{25}{3}$, so $20t-6t^{2}=\frac{25}{3}$. Multiply through by 3 to get $60t - 18t^{2}=25$, or $18t^{2}-60t + 25 = 0$. Using the quadratic formula $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ with $a = 18$, $b=-60$, and $c = 25$, we have $t=\frac{60\pm\sqrt{(-60)^{2}-4\times18\times25}}{2\times18}=\frac{60\pm\sqrt{3600 - 1800}}{36}=\frac{60\pm\sqrt{1800}}{36}=\frac{60\pm30\sqrt{2}}{36}=\frac{10\pm5\sqrt{2}}{6}$ seconds.

Step5: Find time the rock is aloft

The rock is aloft when $s(t)=0$. Set $20t-6t^{2}=0$. Factor out $2t$: $2t(10 - 3t)=0$. So $t = 0$ (initial time) or $10-3t=0$, which gives $t=\frac{10}{3}$ seconds.

Answer:

a) Velocity: $v(t)=20 - 12t$, Acceleration: $a(t)=-12$
b) $\frac{5}{3}$ seconds
c) $\frac{50}{3}$ feet
d) $t=\frac{10 + 5\sqrt{2}}{6}$ seconds and $t=\frac{10 - 5\sqrt{2}}{6}$ seconds
e) $\frac{10}{3}$ seconds