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Applications of Calculus to the physical world (1 Viewer)

Illest

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A rocket is fired vertically from the surface of the earth at a velocity of 5kms^-1. The acceleration of the rocket is given by a= -80000/x^2 kms^-2, where x is the distance of the rocket from the centre of the earth.

If the radius of the Earth is 6400km, find the rocket's velocity when it is 3600km above the Earth's surface.

So I integrated a to get v which is (v^2)/2=80000/x+c

HOW DO I FIND C ?! :(
 

Sy123

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hmmm, I did my way and I got C=0

My answer in the end was:

20/3 km/s

I got the pair by noticing that x is the distance from the earth's surface. Just when the rocket is fired, it is starting to rise at 5km/s however at that point it is 6400km from the centre of the earth. Since C is just a constant we can use this pair to evaluate it

No?
 

Illest

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hmmm, I did my way and I got C=0

My answer in the end was:

20/3 km/s

I got the pair by noticing that x is the distance from the earth's surface. Just when the rocket is fired, it is starting to rise at 5km/s however at that point it is 6400km from the centre of the earth. Since C is just a constant we can use this pair to evaluate it

No?
The answer from the textbook is 4kms^-1
 

RishBonjour

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How can x= 6400 when v=5? & the answer comes out wrong as well.

I thought it would be v=5 when x=0 but there would be no solution.

the x is from the centre of the earth, thus x= 6400, and v = 5.
is the answer different?

i got v = 4 as final answer

EDIT: didnt realise you posted

you must be getting confused with the substitution
it should be

v^2 = 160000/ (6400 +3600)
 

Sy123

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://// Stupid mistake on my part.

Yeah I got 4km/s now. When you find C, you have to subsitute x=10 000 not 3600.

And when you do that, you get 4km/s
 

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