Answer to Question #261818 in Calculus for Dddf

Question

During the first 40 s of a rocket flight, the rocket is propelled straight up so that in t seconds it reaches a height of s=0.3t³ ft. (a) How high does the rocket travel in 40 s? (b) What is the average velocity of the rocket during the first 40 s? (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 40 s?

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

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

(d)

$$v(t)=s'(t)=(0.3t^3)'=0.9t^2$$

$$v(40)=0.9(40)^2$$

$$v(40)=1440ft/s$$