Answer
(a) 19200 ft
(b) 480 ft/s
(c) 66.94 ft/s
(d) 1440 ft/s
Work Step by Step
(a) The height of the rocket in 40s:
$s=0.3t^{3}=0.3\times{40^{3}}=0.3\times64000=19200 ft$
(b) Average speed of the rocket during the first 40s:
Height in 40s = 19200 ft,
Average speed = height/time $= 19200 ft/40s = 480 ft/s$
(c) Average speed of the rocket during the first 1000ft:
Height in 't's = 1000 ft,
$s=0.3t^{3}$
$1000=0.3t^{3}$
$\frac{1000}{0.3}=t^{3}$
$3333.33=t^{3}$
$t=\sqrt[3] 3333.33=14.94s$
Average speed = height/time $= 1000 ft/14.94s = 66.94 ft/s$
(d) The instantaneous velocity (v) of the rocket at the end of 40s:
$v=\frac{ds}{dt}= 0.3\times3t^{2}=0.3\times3(40)^{2}=1440 ft/s$
