Answer to Question #253445 in Calculus for Fahad

Question

During the first 40s of a rocket’s flight, the rocket is propelled straight up so that in t seconds it reaches a height of

s = 0.3t

3

ft.

(a) How high does the rocket travel in 40s?

(b) What is the average velocity of the rocket during the first 40s?

(c) What is the average velocity of the rocket during the first 1000 ft of its flight?

(d) What is the instantaneous velocity of the rocket at the end of 40s?

Answer 1

(a)

$s(40)=0.3(40)^{3}=0.3(64000)=19200 f t$

(b)

$\frac{\Delta s}{\Delta t}=\frac{19200}{40}=480 \mathrm{ft} / \mathrm{s}$

(c)

$\begin{aligned} &1000=0.3 t^{3} \\ &t^{3}=\frac{1000}{0.3}=3333.333 \end{aligned}$

Take the cube root

$t=14.938$ seconds

$\frac{\Delta s}{\Delta t}=\frac{1000}{14.938}=66.94 \mathrm{ft} / \mathrm{s}$

(d)

$\begin{aligned} &\lim _{t \rightarrow 40} \frac{s(40)-s(t)}{40-t} \\ &\lim _{t \rightarrow 40} \frac{19200-0.3 t^{3}}{40-t} \\ &\lim _{t \rightarrow 40} \frac{0.3\left(64000-t^{3}\right)}{40-t} \\ &\lim _{t \rightarrow 40} \frac{0.3(40-t)\left(t^{2}+40 t+1600\right)}{(40-t)} \\ &\lim _{t \rightarrow 40} 0.3\left(t^{2}+40 t+1600\right) \\ &\lim _{t \rightarrow 40} 0.3\left(40^{2}+40(40)+1600\right) \\ &\lim _{t \rightarrow 40} 0.3(4800)=1440 \mathrm{ft} / \mathrm{s} \end{aligned}$