Question

A rocket is shot into the air from the ground so that its height H in feet after t seconds is given by
H(t) = 200t + 10t^2 − 4t^3.

Find the maximum and minimum velocity of the rocket over the interval on which it travels upward.
Find the maximum and minimum velocity of the rocket over the first second of its journey.

Answer 1

do you know how to use derivatives?

Answer 2

are you in calculus? what subject?

Answer 3

@ kimmy0394

Answer 4

can you do derivative of H, d(H)dt ?

Answer 5

yes

Answer 6

ok try that and lets see what you get on dH/dt

Answer 7

f'(x)= 200+20t-12t^2

Answer 8

f''(x)= 20-24t

Answer 9

ok good, that dH/dt = velocity v
for maximum H ,, equate dH/dt=0, and find t, see what you are getting

Answer 10

because max H is attained when v=dH/dt=0

Answer 11

find t there then sub it to H(t) to get max H

Answer 12

so far you did
velocity v=dH/dt=f'(x)= 200+20t-12t^2
now equate it to ZERO,and find t, that is

velocity v=dH/dt=f'(x)= 200+20t-12t^2=0 or

200+20t-12t^2=0, find t?

Answer 13

hint:
use quadratic formula here

Answer 14

\[t=x=\frac{ -b+-\sqrt{b^{2}-4ac} }{ 2a }\]

Answer 15

a=-12 , b=20, c=200 plug them to your quadratic formula and fint t or x

Answer 16

-20+/- sqareroot (-19200) / 24

Answer 17

do the + first then the -

Answer 18

t=(-b+sqrt(b^2 -4ac))/2a then do the negative

t=(-b-sqrt(b^2 -4ac))/2a

Answer 19

watch carefullt you signs specially on negative sign

Answer 20

\[t=\frac{ -20+\sqrt{29^{2}-4(-12)(200)} }{ 2(-12) }\]

Answer 21

29^2? or 20^2?

Answer 22

\[t=\frac{ -20+100 }{ -24 }\]

Answer 23

and
\[t=\frac{ -20-100 }{ -24 }\]

Answer 24

so whats the t1 and t2?

Answer 25

-80/24
and 120/24

Answer 26

yes do it in whole number or mixed number...

Answer 27

so how can you determine max and min

Answer 28

ok good t1=-3.33 sec forget this negative value
and t2= 5 sec ...plug this into you HJ that will be your maxH

Answer 29

ok now that we know how to find the maxH
from the question now, can you find the MAX V?

Answer 30

you plug in 5 to derivative

Answer 31

t= 5sec is to find your maxH,

if you have
velocity V= 200+20t-12t^2
can you try to find the maxV
same as finding maxH

Answer 32

hint find the derivative of V, then equate it to zero and find t then sub it to V
to get the maxV or minV

Answer 33

dV/dt=?
then equate it to zero
dV/dt=0=?

Answer 34

try it first and let me see what u get

Answer 35

from
f'(x)= 200+20t-12t^2 =velocity V
V= 200+20t-12t^2 now do dV/dt=0, fint t? to find the max and min V

Answer 36

from
V= 200+20t-12t^2
is dV/dt=20 - 24t ?

Answer 37

if yes, find
dV/dt=20 -24 t =0, find t=?

Answer 38

is t=20/24 =0.833 sec ?
if yes, sub it to
V= 200+20t-12t^2
V= 200 +20(0.833) - 12(0.833)^2
max V=?

Answer 39

you can use calculator here... lol...:D

Answer 40

is max V=208.33 m/sec ?

Answer 41

my pc froze hold on a sec

Answer 42

contin ue solving them ..lol :D

Answer 43

http://www.wolframalpha.com/input/?i=V%3D+200%2B20t-12t^2

Answer 44

check that out..you can use wolfram on all of your work, for graphing etc

Answer 45

so theres no min value of V here only maxV

Answer 46

Find the maximum and minimum velocity of the rocket over the first second of its journey.

Answer 47

dH/dt= f'(t)=V= 200+20t-12t^2 for the ist sec, t=1 sec sub this to

V= 200+20t-12t^2

Answer 48

is it
for t=1 sec...
max V= 200+20(1)-12(1)^2 = 208 ?