Calculus to the physical world fitzpatrick HELP!!!! (1 Viewer)

[Shikobe]

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Q15. A particle is brought to top speed with an acceleration which varies linearly as the distance traveled. It starts from rest with an acceleration of 3m/s^2 and reaches top speed in a distance of 160m. Find:a) the top speed, b) the speed when the particle has moved 80m.

Q16. A rocket is fired vertically from the earth's surface with an initial speed V. Assuming negligible air resistance Find:a) an expression for v(x) which gives the velocity at a distance x from the centre of the earth, b) how high the rocket will rise, c) the magnitude of V in order that the rocket should escape from the gravitational attraction and never return. Evaluate this particular value of V,given that the radius R of the earth is 6400km.
(Hint: acceleration due to gravity=k/x^2)

OK, this what I've got so far for question 16:

g=k/x^2 (integrate)
v^2/2=-k/x + C
when v=V, x=R
Therefore C=V^2/2+k/x

Therefore, v^2=V^2 - 2k/x + 2k/R

Now, k=gx^2
Therefore, v^2= V^2 - 2gx + 2gx^2/R,

The answer in the book is: v^2=V^2 - 2gR + 2gR^2/x

which is fairly close too what i got up there, but i need too change some x's too R's and some R's too x's. So, WHAT do i do????

Also, part b) i got by putting v=0, and i found x

And part c) i have no clue how too do this. This is escape velocity in physics and the answer is >11200m/s

All helps will be appreciated! HELP ME PLEASE!!!!!!!!!!!!!!

 

[Shikobe]

why isn't anyone replying? Too hard? C'mon...

 

[hscishard]

Haven't learnt that yet.

Biggest help

 

[Shikobe]

Okay i think ive got it now....

a) a=-k/x^2, v^2/2=k/x + C,
when x=R, v=V, therefore C=V^2/2 - k/R

therefore v^2=V^2 + k/x - k/R

Now, g=k/x^2 when x=R( this is what i did wrong)
Therefore, k=gX^2

Now, v^2= V^2 - 2gR + 2gR^2/x