Physics: acceleration questions gah

[dighn]

yes assuming it keeps accelerating backwards even after it stops

 

[acidvoodoo]

ok i sorted the rest of question 5 it wasn't too bad

now i'm on to 6 which is the hardest yet but last

6. A research rocket is fired vertically upwards. It has a uniform acceleration of 400 ms^2 for 12s, after which all it's fuel has been burned. It then travels freely back to the ground. assume the acceleration due to the earths gravity is 10ms^2

a)what is the velocity of the rocket after 12s?
b)At which time does it reach it's maximum height?
c)what is the maximum height?
d)with what velocity does it hit the earth?

a) i'm thinking i just need v=u+at giving v=0+12x400=4800m but now i'm thinking thats quite frickin fast, but oh well it's theoretical.
b)i'm miffed about this. I know that even though it stops being propelled after 12 seconds by the boosters it should keep going until the earths gravitational pull slows it down and it falls back down again, but i can't think how to calculate it and with what equations
c)once again i can calculate how high it got whilst being propelled with s=ut+0.5a(t)^2 toget 28800 metres[ i think] but not sure how to calculate what happens after, any idea?

 

[RossGr]

Originally posted by: acidvoodoo
ok i sorted the rest of question 5 it wasn't too bad

now i'm on to 6 which is the hardest yet but last

6. A research rocket is fired vertically upwards. It has a uniform acceleration of 400 ms^2 for 12s, after which all it's fuel has been burned. It then travels freely back to the ground. assume the acceleration due to the earths gravity is 10ms^2

a)what is the velocity of the rocket after 12s?
b)At which time does it reach it's maximum height?
c)what is the maximum height?
d)with what velocity does it hit the earth?

a) i'm thinking i just need v=u+at giving v=0+12x400=4800m but now i'm thinking thats quite frickin fast, but oh well it's theoretical.

Looks ok, except for the units that should be 4800m/s )it is velocity after all! You need to keep careful track of your units. The units of acceleration * time = m/s^2 * s = m/s

b)i'm miffed about this. I know that even though it stops being propelled after 12 seconds by the boosters it should keep going until the earths gravitational pull slows it down and it falls back down again, but i can't think how to calculate it and with what equations

Use the same equation v = u + at, with u = 4800m/s , a=-10m/s^2 solve for t when v=0 (it has stopped going up)

t = -u/-a = u/a

try to make it a habit of leaving your equation in general terms until you have isolated the factor you are looking for, then plug in your numbers. It is easier to track your units if you do this.

c)once again i can calculate how high it got whilst being propelled with s=ut+0.5a(t)^2 toget 28800 metres[ i think] but not sure how to calculate what happens after, any idea?

Once you get the time to reach the top of travel the distance is easy use

s= .5at^2 + ut + x

let a = -g; u=4800m/s ; x = the height at which the rockets shut off (you should be able to figure this out!)

for the last part, you have its max height so apply the distance equation useing that as the starting height and -g as the acceleration, find the time required to reach the ground (remember your inital velocity will be 0) . Once you have the time to reach the ground the final velocity should be easy.

For future help you may want to check the link in my sig.

 

[acidvoodoo]

Originally posted by: RossGr

Originally posted by: acidvoodoo
ok i sorted the rest of question 5 it wasn't too bad

now i'm on to 6 which is the hardest yet but last

6. A research rocket is fired vertically upwards. It has a uniform acceleration of 400 ms^2 for 12s, after which all it's fuel has been burned. It then travels freely back to the ground. assume the acceleration due to the earths gravity is 10ms^2

a)what is the velocity of the rocket after 12s?
b)At which time does it reach it's maximum height?
c)what is the maximum height?
d)with what velocity does it hit the earth?

a) i'm thinking i just need v=u+at giving v=0+12x400=4800m but now i'm thinking thats quite frickin fast, but oh well it's theoretical.

Looks ok, except for the units that should be 4800m/s )it is velocity after all! You need to keep careful track of your units. The units of acceleration * time = m/s^2 * s = m/s

b)i'm miffed about this. I know that even though it stops being propelled after 12 seconds by the boosters it should keep going until the earths gravitational pull slows it down and it falls back down again, but i can't think how to calculate it and with what equations

Use the same equation v = u + at, with u = 4800m/s , a=-10m/s^2 solve for t when v=0 (it has stopped going up)

t = -u/-a = u/a

try to make it a habit of leaving your equation in general terms until you have isolated the factor you are looking for, then plug in your numbers. It is easier to track your units if you do this.

c)once again i can calculate how high it got whilst being propelled with s=ut+0.5a(t)^2 toget 28800 metres[ i think] but not sure how to calculate what happens after, any idea?

Once you get the time to reach the top of travel the distance is easy use

s= .5at^2 + ut + x

let a = -g; u=4800m/s ; x = the height at which the rockets shut off (you should be able to figure this out!)

for the last part, you have its max height so apply the distance equation useing that as the starting height and -g as the acceleration, find the time required to reach the ground (remember your inital velocity will be 0) . Once you have the time to reach the ground the final velocity should be easy.

For future help you may want to check the link in my sig.

sorry isn't it t=-u/a but then as a is negative it still works?

 

[RossGr]

notice that in my original expression I wrote

t = -u/-a
The 2 negitives cancel.

 

[acidvoodoo]

yea but when i re arranged the initial equation i did

v=u+at
0=u+at
-u=at
-u/a=t

is it just that you knew the a value is going to be negative anyway so you put a minus sign on it first to make the equation all positive. ahhhhhhh i'm confused but i got an answer of 480 seconds for that part

 

[RossGr]

Yes,you are correct, that was bad form on my part. Sorry!

 

[acidvoodoo]

for max height i got

s=4800x12 +0.5x-10x(12)^2 giving 56880m does this look correct?

 

[acidvoodoo]

and then for the last part i got


56880=0xt+ .5x-10x(t)^2
56880=.5x-10x(t)^2
113760=-10x(t)^2
113760/-10=-11376

-11376=t^2 root both sides and my calc gives 106.658333 I <-----------no idea what this i means, i know in math you can't suare root negative numbers but in physics it's a direction i think so it doesn't matter, but should i write down the I

 

[acidvoodoo]

ooo er i just realised about that x, where did that come from in yer equation s=.5a(t)^2+ut+x

 

[ebaycj]

Originally posted by: Chaotic42
Wouldn't you just use the acceleration formula?

s=(vo)t + 1/2at²

I don't know about that m/s^-2.

I've always seen it as m/s^2.

yeah, what he said before was not m/s^2, it was (m)(s^-2). they are the same thing.

 

[RossGr]

Originally posted by: acidvoodoo
ooo er i just realised about that x, where did that come from in yer equation s=.5a(t)^2+ut+x

The x is your initial position.

 

[acidvoodoo]

ok so it should look like


s=0.5*-10*(t)^2+4800*t+56880


i must have done it wrong cause ther eis 2 unknowns?

 

[RossGr]

Originally posted by: acidvoodoo
ok so it should look like


s=0.5*-10*(t)^2+4800*t+56880


i must have done it wrong cause ther eis 2 unknowns?

You have been given a lot of very explicit help in this thread, it is about time for you to break free and fly on your own. You have 2 basic relationships.
1. v = u + at

where v is your final position and u is your initial velocity.

2. s = .5 a t^2 + ut + x

where x is your initial position and u is still initial velocity.

You must examine your problem and determine what you need to find and what you are given. Use that to determine which equation you must use.

So for your last post,

what are you looing for?

What do you have?

 

[acidvoodoo]

56880=0xt+ .5x-10x(t)^2
56880=.5x-10x(t)^2
113760=-10x(t)^2
113760/-10=-11376

-11376=t^2 root both sides and my calc gives 106.658333 with an i next to it, so thats what i got for the time

for velocity i get v=u+at

=0+-10*106.658
=-1066.58m/s

does that look ok?

 

[RossGr]

you are very close, I would set it up with x=your max height and then find t when s=0 (you reach the ground) That should cure your sign problem.

 

[acidvoodoo]

so now that the signs are sorted out is it the correct answer? like is my second part correct?

 

[silverpig]

Um, are we assuming that 400m*sec^-2 = it's acceleration after taking gravity into account?

Seems to me as though you would have an acceleration of 390 m*sec^-2 for 12 seconds, although it depends on what your teacher expects. It doesn't specify in the question though.

 

[Tweak155]

m*s^-2 is m/s/s (meters per second per second) can also be written as:

m/(s^2), m/(s*s)....anything to a negative power is put in the denominator