on another planet, a rock is thrown upward with a velocity of 19\\(\\frac{m}{s}\\). the rocks height in…

Explanation:

Step1: Recall the velocity formula

The velocity (v(t)) is the derivative of the height function (h(t)). Using the power rule ((x^n)^\prime=nx^{n - 1}), for (h(t)=19t-3.101t^{2}), we have (v(t)=h^\prime(t)). [ \begin{align*} v(t)&=\frac{d}{dt}(19t)-\frac{d}{dt}(3.101t^{2})\ \end{align*} ]

Step2: Differentiate each term

For the first - term (\frac{d}{dt}(19t)), since (\frac{d}{dt}(ax)=a) (where (a = 19)), (\frac{d}{dt}(19t)=19). For the second - term (\frac{d}{dt}(3.101t^{2})), using the power rule (\frac{d}{dt}(ax^{n})=nax^{n - 1}) with (a = 3.101) and (n = 2), we get (\frac{d}{dt}(3.101t^{2})=2\times3.101t=6.202t). So (v(t)=19 - 6.202t).

Part a)

Substitute (t = a) into (v(t)): (v(a)=19-6.202a)

Part b)

Substitute (t = 1.8) into (v(t)): (v(1.8)=19-6.202\times1.8) [ \begin{align*} v(1.8)&=19-(6.202\times1.8)\ &=19 - 11.1636\ &=7.8364 \end{align*} ]

Part c)

The rock hits the ground when (h(t)=0). So (19t-3.101t^{2}=t(19 - 3.101t)=0). We have two solutions: (t = 0) (initial time) and (19-3.101t=0). Solving (19-3.101t=0) for (t): [ \begin{align*} 3.101t&=19\ t&=\frac{19}{3.101}\approx6.1 \end{align*} ]

Part d)

Substitute (t=\frac{19}{3.101}) into (v(t)): [ \begin{align*} v\left(\frac{19}{3.101}\right)&=19-6.202\times\frac{19}{3.101}\ &=19-(2\times19)\ &=19 - 38\ &=- 19 \end{align*} ]

Answer:

a) (19 - 6.202a) b) (7.8364) c) (6.1) d) (-19)