Physical world problems (1 Viewer)

[catmaster]

Can someone please help me with these questions:

A particle brought to top speed with an acceleration which varies linearly (a straight line?) as the distance travelled. Its starts from rest with an acceleration of 3m/s^2 and reaches top speed in a distance of 160 metres. Find: (a)The top speed, (b) the speed when the particle has moved 80metres.
I am speculating whether the questions does not provide enough information or I misunderstood the question....what does linearly mean.....I know what linear functions are but what does 'varies linearly' mean

A rocket is fired vertically from the earth's surface with an initial speed v. Assuming negligable air resistance: (a) Find how high the rocket will rise, (b) find the magnitude of v in order that the rocket should escape from the gravitational attraction and never return. Evaluate this value of v, given radius of earth is 6400 km
I managed to get part of a out by my final answer differed from the textbook's. I did not bother with b since i did not fully solve (a)...and this questions sounds awfully like the concept of 'escape velocity' from physics.....

Ummmm
here is my working from the second question part a

let:
gravity=9.8=g
accelleration = A
Velocity = V
initial velocity = v (small v)
Displacement= X
Radius of earth = r

Now
A = g
but since A = d(half V squared)/dx
therefore (half V squared) = integrate g
(half V squared)= gx + c
(V squared) = 2gx + 2c....................................(@)
but at x=r
V=v (small v)
therefore
(small v squared) = 2gr +2c
2c = (small v squared) - 2gr
[sub this back into (@)]
(Big V squared) = 2gx + (small v squared) - 2gr
...........which is not equal to the answers
answers:
(Big V squared) = 2g((r squared)/x) + (small v squared) - 2gr

I am really lost....my working seems logical ......well to me it does


(these questions are from New Senior Mathematics yr 12 3unit chapter 25(b) the third and second last questions..i.e. Q15 and Q16)


If anyone would help that would be wonderful!! thank you
(I hate textbooks which don't gived worked solutions....very few do.....like terry lee's textbook.....soo...expensive......)

 

[vds700]

15) a = 3 + kx
when x = 160, a = 0
3 + 160k = 0
k = -3/160

(1/2)v^2 = I (3 + kx)dx
= 3x + kx^2/2 + c when x = 0, v = 0, therfore c = 0
v^2 = 6x + kx^2
= 6x - 3x^2/160

a)when x = 160,
v^2 = 6(160) - 3(160)
= 480
theerfore v = 21.91 m/s -> max apeed

b)when x = 80,
v^2 = 6(80) - 3(80)^2/160
=360
therefore v = 18.97 m/s

the acceleratin varies linearly, so i think it means its in the form of a straight line graph, y = kx + c, but when x=0, a = 3, so c = 3

16)i think the answer may be wrong to this question, i cant get it and my teacher couldn't get it. It says underneath the question that a = k/x^2, but the answer has g in it, where did fitzy get the g from?

 

[Iruka]

The acceleration is not a constant, its a function of x. (inverse square of the distance).

So a = k/x^2, but when x=r, a = -g (since the gravitational acceleration is down).

Hence k = -gr^2

 

[vds700]

The acceleration is not a constant, its a function of x. (inverse square of the distance).

So a = k/x^2, but when x=r, a = -g (since the gravitational acceleration is down).

Hence k = -gr^2

ah got it now
a = -gR^2/x^2
a)find v(x)
(1/2)v^2 = -gR^2 . I x^-2 dx
= -gR^2 (-1/x) + c
=gR^2/x + c
when v = V, x = R, c = (1/2)V^2 - gR
(1/2)v^2 = gR^2 /x + (1/2)V^2 - gR
v^2 = 2gR^2/x + V^2 - 2gR

b)rearrange , make v = 0 to get x = 2gR^2/(2gR - V^2)

c)to escape the earth's gravitatinal attraction, x = infinity
so 2gR - V^2 = 0
V^2 = 2gR
v=sqrt(2 x 9.8 x 6400 x 1000)
= 11200 m/s

 

[cherichan]

ah got it now
a = -gR^2/x^2
a)find v(x)
(1/2)v^2 = -gR^2 . I x^-2 dx
= -gR^2 (-1/x) + c
=gR^2/x + c
when v = V, x = R, c = (1/2)V^2 - gR
(1/2)v^2 = gR^2 /x + (1/2)V^2 - gR
v^2 = 2gR^2/x + V^2 - 2gR

b)rearrange , make v = 0 to get x = 2gR^2/(2gR - V^2)

c)to escape the earth's gravitatinal attraction, x = infinity
so 2gR - V^2 = 0
V^2 = 2gR
v=sqrt(2 x 9.8 x 6400 x 1000)
= 11200 m/s

Just a little question: for part c, why did you multiple it by 1000?
Doesn't V^2 = 2gR
therefore V=sqrt(2x9.8x6400)?

Thanks so much for the solution

 

[RishBonjour]

Just a little question: for part c, why did you multiple it by 1000?
Doesn't V^2 = 2gR
therefore V=sqrt(2x9.8x6400)?

Thanks so much for the solution

the answer is in m/s but its 6400km so you multiply by 1000
fml. been stuck on this question for a while!