Physics: acceleration questions gah

[acidvoodoo]

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

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? to get 750m but i don't think this is right?


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 106.658333s 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?

 

[Chaotic42]

5 meters per second ^ -2?

What the hell? 1/sec²

That's weird.

 

[acidvoodoo]

yes this is the type of notation we use at my school, i don't really get it either

for acceleration it uses to the power -2 and for velocity it uses to the power -1 , maybe it's different here in england

 

[StageLeft]

You know it starts at a zero speed. After 30 seconds it's going at 30 X whatever the acceleration was right? Thus, its average speed is exactly half of that. So then multiply its average speed times 30 seconds and that's the answer!

 

[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.

 

[acidvoodoo]

so i got 2250m by doing that, is that correct?

edit oh ok i get it, supposed to use that equation

 

[StageLeft]

yeah there is a formula, but I Just forgot what it was :0

I get 30 seconds x 5ms gives you a final velocity of 150 m/s. Your average velocity was 75 m/s, so 75 ms X 30 seconds = 2250 m. That's a damned fast train

 

[Chaotic42]

Originally posted by: Skoorb
yeah there is a formula, but I Just forgot what it was :0

I get 30 seconds x 5ms gives you a final velocity of 150 m/s. Your average velocity was 75 m/s, so 75 ms X 30 seconds = 2250 m. That's a damned fast train

That's also what I get.

 

[acidvoodoo]

what about uniform deceleration.

"if a motorbike is travelling for 90 seconds at a constant speed of 20m/s and then slows down with a uniform deceleration for 4 seconds until it comes to a rest how far has it travelled"

now i know that the first part of the question is 1800m, and with the last part i can work it out by knowing that it obviously decelerated at 5m every second, so i can work out that it travelled 15m for 1 second, 10m for the next second, 5m in the 3rd second, and by the 4th second it is stopped, meaning that is travelled an extra 30m whilst slowing down [i think my logic is correct anyway]. so that gives a distance of 1830m The problem is, i don't think that is the proper way of working it out(the last part), and i'm not sure how to do it the proper way [this homework has alot of marks coming from the working out that's why i'm so worried about getting it right]

do either of you guys have a different method of working out the decelerating part

 

[StageLeft]

Originally posted by: acidvoodoo
what about uniform deceleration.

"if a motorbike is travelling for 90 seconds at a constant speed of 20m/s and then slows down with a uniform deceleration for 4 seconds until it comes to a rest how far has it travelled"

now i know that the first part of the question is 1800m, and with the last part i can work it out by knowing that it obviously decelerated at 5m every second, so i can work out that it travelled 15m for 1 second, 10m for the next second, 5m in the 3rd second, and by the 4th second it is stopped, meaning that is travelled an extra 30m whilst slowing down [i think my logic is correct anyway]. so that gives a distance of 1830m The problem is, i don't think that is the proper way of working it out(the last part), and i'm not sure how to do it the proper way [this homework has alot of marks coming from the working out that's why i'm so worried about getting it right]

do either of you guys have a different method of working out the decelerating part

Well you're right to break down the solution into two parts. The first part, which gives you a distance of 1800 m, then the second part.

The second part of the question is really the same as the train question except in the other direction (deceleration instead of acceleration). In this case it starts at 20m/s speed and decreases down to 0 m/s speed. Since uniform deceleration will give it an average speed during this braking period of 10 m/s (figured by [starting speed + end speed]/2 ), you know that it travelled at an average speed of 10m/s, for 4 seconds = 40 m. So the final answer is 1840.

The reason your second by second analysis gives the wrong answer is because from 0-1 seconds it's not travelling at 20 or at 15, but rather at 17.5....so 17.5 + 12.5 (for seconds 1-2), + 7.5 (2-3), and finally + 2.5 (seconds 3-4) will give you the 40

 

[acidvoodoo]

ohhhhhh i get it now, this is one of those moments where i was looking for a complicated solution

would you [or anyone] mind checking my answers to this question

4- A ball takes 450ms to fall 1.0m [so thats milliseconds i'm sure]

a)what value does this give for the acceleration due to gravity?
b)how long would the ball take to fall 25cm?
c)if it continued with the same acceleration, how long would it take in falling from a cliff 100m high?
d)In fact, it would not continue with the same acceleration for 100m, why not?

for question a I used the equation v=at to get 9.8 x 0.45 [9.8 is the acceleration of free fall on earth and it didn't specify that it wasn't on earth] so i got a velocity of 4.41 m/s

for question b i wasn't completly sure, it's falling 4 times less of a distance so i divded the 0.45ms by 4 to get 0.1125ms but i'm not sure if that is the correct method

for question c i took the equation
s=ut+0.5a(t)^2
100=0xt+0.5x9.8(t)^2
re arranged to get

100/(0.5x9.8)=t^2
100/4.9=t^2
20.401=t^2
square rooted both sides
got t=4.517 seconds

and for question d i thought that the velocity would reach 9.8m/s and not be able to accelerate any longer but i'm not sure about this

 

[acidvoodoo]

bump

 

[dighn]

Originally posted by: acidvoodoo
ohhhhhh i get it now, this is one of those moments where i was looking for a complicated solution

would you [or anyone] mind checking my answers to this question

4- A ball takes 450ms to fall 1.0m [so thats milliseconds i'm sure]

a)what value does this give for the acceleration due to gravity?
b)how long would the ball take to fall 25cm?
c)if it continued with the same acceleration, how long would it take in falling from a cliff 100m high?
d)In fact, it would not continue with the same acceleration for 100m, why not?

for question a I used the equation v=at to get 9.8 x 0.45 [9.8 is the acceleration of free fall on earth and it didn't specify that it wasn't on earth] so i got a velocity of 4.41 m/s

for question b i wasn't completly sure, it's falling 4 times less of a distance so i divded the 0.45ms by 4 to get 0.1125ms but i'm not sure if that is the correct method

for question c i took the equation
s=ut+0.5a(t)^2
100=0xt+0.5x9.8(t)^2
re arranged to get

100/(0.5x9.8)=t^2
100/4.9=t^2
20.401=t^2
square rooted both sides
got t=4.517 seconds

and for question d i thought that the velocity would reach 9.8m/s and not be able to accelerate any longer but i'm not sure about this

a) not right. use s = ut + 0.5at^2 (i would assume 0 initial v.) just plugin the values
b) freefall is nonlinear with respect to time. you can't simply divide by 4. use s = ut + 0.5at^2. plugin values
c) your method is right. except your value of a from a) is wrong
d) i can think of 2 reasons: air friction and the closer you get to the earth the more the acceleration. the latter should be negliable though for 100 meters

 

[esun]

4- A ball takes 450ms to fall 1.0m [so thats milliseconds i'm sure]

a)what value does this give for the acceleration due to gravity?
b)how long would the ball take to fall 25cm?
c)if it continued with the same acceleration, how long would it take in falling from a cliff 100m high?
d)In fact, it would not continue with the same acceleration for 100m, why not?

for question a I used the equation v=at to get 9.8 x 0.45 [9.8 is the acceleration of free fall on earth and it didn't specify that it wasn't on earth] so i got a velocity of 4.41 m/s

This doesn't answer the question (it asks for acceleration due to gravity, not velocity). You'll need to use x = 1/2at^2, plug in x = 1, t = 0.450, then solve for a.

for question b i wasn't completly sure, it's falling 4 times less of a distance so i divded the 0.45ms by 4 to get 0.1125ms but i'm not sure if that is the correct method

x = 1/2at^2. Now that you know a (from part a), plug in x = 0.25 m, a from part a, then solve for t.

for question c i took the equation
s=ut+0.5a(t)^2
100=0xt+0.5x9.8(t)^2
re arranged to get

100/(0.5x9.8)=t^2
100/4.9=t^2
20.401=t^2
square rooted both sides
got t=4.517 seconds

You should do yf = y0 + 1/2at^2 (assuming it starts with v0 = 0). So you know y0 = 100 m, you know a (from part a, but make sure it is negative here because it is falling down!), and you yf = 0 m (it hits the ground). So again, plug in and solve for t.

For part d, there are two reasons that come to mind: first, if we assume this is acceleration due to gravity, then we have that it is getting closer to the center of the earth (or wherever it is falling), therefore the F = k/r^2 relationship is getting slightly larger as r is getting slightly smaller. The more important reason is probably air resistance (assuming where it is falling has an atmosphere).

 

[acidvoodoo]

oh right, i get it, so basicly i just worked out the speed of gravity from the data given [in reply to one of the above people]

 

[acidvoodoo]

Originally posted by: esun
4- A ball takes 450ms to fall 1.0m [so thats milliseconds i'm sure]



You should do yf = y0 + 1/2at^2 (assuming it starts with v0 = 0). So you know y0 = 100 m, you know a (from part a, but make sure it is negative here because it is falling down!), and you yf = 0 m (it hits the ground). So again, plug in and solve for t.

For part d, there are two reasons that come to mind: first, if we assume this is acceleration due to gravity, then we have that it is getting closer to the center of the earth (or wherever it is falling), therefore the F = k/r^2 relationship is getting slightly larger as r is getting slightly smaller. The more important reason is probably air resistance (assuming where it is falling has an atmosphere).

why must the acceleration be negative for c and not the other questions. and also, i've run into a problem because i end up with a negative number being equal to t^2, so i can't square root both sides [on question c, maybe i did it wrong, i used the equation s=ut+1/2at^2], however if i keep it positive and can root both sides iget 4.5 seconds

 

[esun]

Well technically it should be negative for all parts, as should the velocity, since it is falling downward. I just want to make sure that in case you omitted it previously, you include it in part c. This is because the way I set up the equation, zero is the ground, and 100 m cliff is positive, meaning upwards. That means if it is accelerating down, it must be accelerating in the negative direction.

Another way to set it up would be to consider zero as the top of the cliff and downward the positive direction, so your equation would look like this:

100 = 0 + 1/2at^2

now with a positive (y0 = 0, yf = 100 m down, but since a is pointing in that same direction, it is positive).

EDIT:

Also, that equation is wrong (s = ut + 1/2at^2). I'm assuming u is velocity here, and the initial velocity is zero (assuming it starts with zero velocity at the top of the cliff, which I believe is implied). However, it does have an initial position that is relevant, which is why you need y0 (or s0 in your notation), so s = s0 + 1/2at^2 is the right equation.

 

[acidvoodoo]

well i did it the second way. would both ways be allowed, all of these questions were quite vague in terms on things like that and how many decimal places to put for answers etc.

ok i'm done with question 4 do you haven any advice for question 5? i have no idea where to start [it's in my first post]

 

[dighn]

Originally posted by: acidvoodoo
well i did it the second way. would both ways be allowed, all of these questions were quite vague in terms on things like that and how many decimal places to put for answers etc.

ok i'm done with question 4 do you haven any advice for question 5? i have no idea where to start [it's in my first post]

you talking about this one: ?

A tennis ball, moving to the right with a velocity of 25m/s, starts to decelerate unifromly at a rate of 200ms^2 when t =0.00s

a)calculate the time taken for the ball to reach zero velocity

i assume gravity is not involved here. anyway. delta v = a*delta t. delta v = -25, a = -200, t = 25/200 = 0.125

 

[acidvoodoo]

sorry i do not understand that notation 3 posts up

 

[dighn]

.

 

[acidvoodoo]

Originally posted by: dighn

Originally posted by: acidvoodoo
well i did it the second way. would both ways be allowed, all of these questions were quite vague in terms on things like that and how many decimal places to put for answers etc.

ok i'm done with question 4 do you haven any advice for question 5? i have no idea where to start [it's in my first post]

you talking about this one: ?

A tennis ball, moving to the right with a velocity of 25m/s, starts to decelerate unifromly at a rate of 200ms^2 when t =0.00s

a)calculate the time taken for the ball to reach zero velocity

i assume gravity is not involved here. anyway. delta v = a*delta t. delta v = -25, a = -200, t = 25/200 = 0.125

why's the velocity negative here [not that i don't listen in class the teacher just likes to to be challenged ]

 

[WinkOsmosis]

What? That tennis ball problem is easy. 25/200 = 1/8 second.

 

[dighn]

Originally posted by: acidvoodoo

Originally posted by: dighn

Originally posted by: acidvoodoo
well i did it the second way. would both ways be allowed, all of these questions were quite vague in terms on things like that and how many decimal places to put for answers etc.

ok i'm done with question 4 do you haven any advice for question 5? i have no idea where to start [it's in my first post]

you talking about this one: ?

A tennis ball, moving to the right with a velocity of 25m/s, starts to decelerate unifromly at a rate of 200ms^2 when t =0.00s

a)calculate the time taken for the ball to reach zero velocity

i assume gravity is not involved here. anyway. delta v = a*delta t. delta v = -25, a = -200, t = 25/200 = 0.125

why's the velocity negative here [not that i don't listen in class the teacher just likes to to be challenged ]

delta v = Vfinal - Vinitial

 

[acidvoodoo]

Originally posted by: dighn

Originally posted by: acidvoodoo

Originally posted by: dighn

Originally posted by: acidvoodoo
well i did it the second way. would both ways be allowed, all of these questions were quite vague in terms on things like that and how many decimal places to put for answers etc.

ok i'm done with question 4 do you haven any advice for question 5? i have no idea where to start [it's in my first post]

you talking about this one: ?

A tennis ball, moving to the right with a velocity of 25m/s, starts to decelerate unifromly at a rate of 200ms^2 when t =0.00s

a)calculate the time taken for the ball to reach zero velocity

i assume gravity is not involved here. anyway. delta v = a*delta t. delta v = -25, a = -200, t = 25/200 = 0.125

why's the velocity negative here [not that i don't listen in class the teacher just likes to to be challenged ]

delta v = Vfinal - Vinitial

oh i see so it was 0-25=-25


ok for part b of qestion 5 it asks "calculate the velocity of the ball after 0.15s"

i used v=u+at to get
v=25+-200x0.15
v=25+-30
v=-5ms right?