Question

if a rock is thrown upward on the planet mars with a velocity of 13 m/s, its height above the ground (in meters) after t seconds is given by h = 13t - 1.86t². (a) find the velocity (in m/s) of the rock after 2 seconds. (b) find the velocity (in m/s) of the rock when t = a. (c) when (in seconds) will the rock hit the surface? (round your answer to one decimal place.) (d) with what velocity (in m/s) will the rock hit the surface?

Explanation:

Step1: Recall the velocity - height relationship

The velocity $v(t)$ is the derivative of the height function $H(t)$. Given $H(t)=13t - 1.86t^{2}$, using the power - rule $\frac{d}{dt}(at^{n})=nat^{n - 1}$, we have $v(t)=H^\prime(t)=13-3.72t$.

Step2: Solve part (a)

Substitute $t = 2$ into the velocity function $v(t)$. So $v(2)=13-3.72\times2=13 - 7.44 = 5.56$ m/s.

Step3: Solve part (b)

Substitute $t = a$ into the velocity function $v(t)$. So $v(a)=13-3.72a$ m/s.

Step4: Solve part (c)

The rock hits the surface when $H(t)=0$. So we set $13t-1.86t^{2}=0$. Factor out $t$: $t(13 - 1.86t)=0$. We have two solutions: $t = 0$ (corresponds to the initial time) and $13-1.86t=0$. Solving $13-1.86t=0$ for $t$, we get $t=\frac{13}{1.86}\approx7.0$ s.

Step5: Solve part (d)

Substitute the time $t=\frac{13}{1.86}$ into the velocity function $v(t)$. $v(\frac{13}{1.86})=13-3.72\times\frac{13}{1.86}=13 - 26=- 13$ m/s.

Answer:

(a) $5.56$ m/s
(b) $13 - 3.72a$ m/s
(c) $7.0$ s
(d) $-13$ m/s