Flight of the Rocket
A rockets launch site is 2000 ft. from the control center. The rocket blasts off vertically at the rate of 500 ft./sec. After 10 seconds, how fast is the distance changing between the rocket and the control center?
posted by Zaux at
9:47 PM
6 Comments:
khalid
maryam
nah ... I believe it's William Shakespeare ... whew! ... must be some good stuff!
Flight of the Rocket
Given: Launch site is 2000’ from rocket
Vertical rate of climb of rocket: 500’/s
dD/dt=dD/dx*dx/dt
distance from control centre to rocket is:
D=(x^2+y^2)^0.5
dD/dx=0.5*(x^2+y^2)^-0.5*2*x
dD/dx=x/sqrt(x^2+y^2)
at 10s x=500*10=5000
dD/dx=500/sqrt(5000^2+2000^2)~=0.092848
dx/dt=500
dD/dt~=0.092848*500~=46.42
Answer:
~46.42 ft/s
Cam
Hi Cam you made a small error (prolly a typo).
When you calculated dD/dx you accidentally used 500 instead of 5000 for x in the numerator.
You should have ended up with 464.2 fps.
Oops,I got a bit sloppy there. Thanks Chris. The corrected solution is:
Flight of the Rocket
Given: Launch site is 2000’ from rocket
Vertical rate of climb of rocket: 500’/s
dD/dt=dD/dx*dx/dt
distance from control centre to rocket is:
D=(x^2+y^2)^0.5
dD/dx=0.5*(x^2+y^2)^-0.5*2*x
dD/dx=x/sqrt(x^2+y^2)
at 10s x=500*10=5000
dD/dx=5000/sqrt(5000^2+2000^2)~=0.92848
dx/dt=500
dD/dt~=0.92848*5000~=464.2
Answer:
~464.2 ft/s
Cam
Chris ... you beat me to it ... was gonna post --> right digits, wrong decimal point place
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