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Atheism
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9/16/2012 7:14:27 PM
Posted: 4 years ago
This problem was on a test of my friend's, and she would like to know how to do it, someone to solve it, etc. because she's worried about her grade. I'm pretty poor/lazy at physics, so if someone could help, that'd be awesome.

A car accelerates for 4 seconds, then maintains a constant velocity for 14 seconds. At the end of 18 seconds, it has traveled 1200 m.
What is a) Its initial acceleration, and
b) the distance at which it stopped accelerating?

Now assume that the car decelerates at -12.5 m/s.
How long does it take for it to come to a full stop, and at what distance?
I miss the old members.
Ren
Posts: 7,102
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9/16/2012 7:50:56 PM
Posted: 4 years ago
At 9/16/2012 7:14:27 PM, Atheism wrote:
This problem was on a test of my friend's, and she would like to know how to do it, someone to solve it, etc. because she's worried about her grade. I'm pretty poor/lazy at physics, so if someone could help, that'd be awesome.

A car accelerates for 4 seconds, then maintains a constant velocity for 14 seconds. At the end of 18 seconds, it has traveled 1200 m.
What is a) Its initial acceleration, and
b) the distance at which it stopped accelerating?

Now assume that the car decelerates at -12.5 m/s.
How long does it take for it to come to a full stop, and at what distance?

Welllllll, I have to be honest, I don't actually know the formula for this; I don't remember. So, at times like this, I'd try to figure it out logically. This is what I did.

Let's first divide 1,200 by 18 to see what the speed would be if the car were moving at a constant velocity the entire time. We end up with 66.667.

Now, we also know that one way or another, the car never reached 100 m/s. We know that, because it would have exceed 1,200 meters within those last 14 seconds alone. So, more than 66.667, and less than 100. Let's try 75.

14 * 75 is 1,050.

Now, let's add in the first four seconds of acceleration. If the highest speed reached is 75 m/s with 4 seconds to reach that velocity, then the acceleration would be 25 m/s.

1 - 0
2 - 25
3 - 50
4 - 75
= 150 + 1,050 = 1,200

So, the answer to a) is 25 m/s, and the answer to b) is 150 m.

The second question is much easier. At -12.5 m/s acceleration, you're looking at 1,200/12.5 = 96.

So, it would take 96 seconds to decelerate to a full stop.

The distance you'd travel is a little more tricky, though. Let's reduce the numbers down, so it's not as retarded. It would take 9.6 seconds to decelerate from 120 m/s to 0 at a rate of 12.5 m/s.

1. 120 - 12.5
2. 107.5 - 12.5
3. 95 - 12.5
4. 82.5 - 12.5
5. 70 - 12.5
6. 57.5 - 12.5
7. 45 - 12.5
8. 32.5 - 12.5
9. 20 - 12.5
0.6. 7.5 - 7.5 (12.5 * 0.6 = 7.5) = 0

120 + 107.5 + 95 + 82.5 + 70 + 57.5 + 45 + 32.5 + 20 + 7.5 = 637.5

Scale that back up, and we're at 6,375 m.

Lol, I know I'm not doing this the "official" way (there's undoubtedly algorithms and specific equations one should utilize), but this is still a viable way to, if nothing else, check your answers. :P

I'm just too lazy to look it up right now; I honestly don't remember the official way to do it.
Ren
Posts: 7,102
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9/16/2012 10:07:23 PM
Posted: 4 years ago
Well, coming back to this, I wonder why I started the first second at a stop... but, I guess it's best to always look over your answers to make sure you did them right.

I initially stated:

1 - 0
2 - 25
3 - 50
4 - 75
= 150 + 1,050 = 1,200

So, the answer to a) is 25 m/s, and the answer to b) is 150 m.

Which is completely wrong... it would instead be:

1 - 18.75 +
2 - 37.5 +
3 - 56.25 +
4 - 75 =

187.5 + (75 * 14) = 1237.5

...which is wrong, so it couldn't be 75. I guess I was just figuring it in such a way that it would make sense.

So, I think I'm going to try this on paper, instead, before I go looking for the acceleration transformation that applies to this (you're not left with much information).

...

...well, the direction I ended up going was to reduce it to its smallest increment algebraically and work from there.

What we have here is the initial acceleration applied four times in a stepwise fashion, then that acceleration four times over 14 times repeated... lol... which translates in math terms as:

(x + 2x + 3x + 4x) + 14(4x) = 1,200.
(10x) + 14(4x) = 1,200
10x + 56x = 1,200
66x = 1,200
x = 1,200/66 = 18.1818182

18.1818182 * 4 = 72.7272728

So, in those first four seconds, it was accelerating at 18.1818182 m/s, and it travelled 181.8181819 m by the end of those first four seconds. It thus traveled 72.7272728 m/s thereafter to finally reach 1,200 m at the end of 18 s.

Hilariously, it works out.
darkkermit
Posts: 11,204
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9/16/2012 10:35:23 PM
Posted: 4 years ago
At 9/16/2012 7:50:56 PM, Ren wrote:
At 9/16/2012 7:14:27 PM, Atheism wrote:
This problem was on a test of my friend's, and she would like to know how to do it, someone to solve it, etc. because she's worried about her grade. I'm pretty poor/lazy at physics, so if someone could help, that'd be awesome.

A car accelerates for 4 seconds, then maintains a constant velocity for 14 seconds. At the end of 18 seconds, it has traveled 1200 m.
What is a) Its initial acceleration, and
b) the distance at which it stopped accelerating?

Now assume that the car decelerates at -12.5 m/s.
How long does it take for it to come to a full stop, and at what distance?

Welllllll, I have to be honest, I don't actually know the formula for this; I don't remember. So, at times like this, I'd try to figure it out logically. This is what I did.

Let's first divide 1,200 by 18 to see what the speed would be if the car were moving at a constant velocity the entire time. We end up with 66.667.

Now, we also know that one way or another, the car never reached 100 m/s. We know that, because it would have exceed 1,200 meters within those last 14 seconds alone. So, more than 66.667, and less than 100. Let's try 75.

14 * 75 is 1,050.

Now, let's add in the first four seconds of acceleration. If the highest speed reached is 75 m/s with 4 seconds to reach that velocity, then the acceleration would be 25 m/s.

1 - 0
2 - 25
3 - 50
4 - 75
= 150 + 1,050 = 1,200

So, the answer to a) is 25 m/s, and the answer to b) is 150 m.

The second question is much easier. At -12.5 m/s acceleration, you're looking at 1,200/12.5 = 96.

So, it would take 96 seconds to decelerate to a full stop.

The distance you'd travel is a little more tricky, though. Let's reduce the numbers down, so it's not as retarded. It would take 9.6 seconds to decelerate from 120 m/s to 0 at a rate of 12.5 m/s.

1. 120 - 12.5
2. 107.5 - 12.5
3. 95 - 12.5
4. 82.5 - 12.5
5. 70 - 12.5
6. 57.5 - 12.5
7. 45 - 12.5
8. 32.5 - 12.5
9. 20 - 12.5
0.6. 7.5 - 7.5 (12.5 * 0.6 = 7.5) = 0

120 + 107.5 + 95 + 82.5 + 70 + 57.5 + 45 + 32.5 + 20 + 7.5 = 637.5

Scale that back up, and we're at 6,375 m.

Lol, I know I'm not doing this the "official" way (there's undoubtedly algorithms and specific equations one should utilize), but this is still a viable way to, if nothing else, check your answers. :P

I'm just too lazy to look it up right now; I honestly don't remember the official way to do it.

wrong answer is wrong.

The actually equation your looking for is this

d = .5(A)*t^2 + (vo)*t

d = distance, Ao = acceleration, vo = initial velocity

However it should be noted that the above equation assumes a constant acceleration. In the problem, in the first 4 seconds it has a constant acceleration, and in the next 14 seconds, it has a constant velocity.

So
d1 = .5(A)(t1)^2 [One can ignore the initial velocity, since an initial velocity of 0 is assumed.t1 = time, d1 = distance]

d2 = Vo(t2) [One can ignore the acceleration portion, since acceleartion is 0]

So right now there are 4 unknowns, and only 2 equations (for every unknown you need one equation).

A third equation is needed.
d1 + d2 = 1200. We know that the sum of the two distances are 1200 meters from the problems.

The forth equation is as follows:

v0 = A*t1. Acceleration*time = velocity.

So using some substitution one arrives at one giant equation:

d1 + d2 = .5(A)(t1)^2 + Vo(t2) = 1200

d1 + d2 = .5(A)(t1)^2 + A*t1*(t2) = 1200
.5*A*4^2 + A*14*4 = 1200

Using basic algebra
A = acceleration = 18.75

Also find vo, since this will be important for the final problem
A*t1.= 18.75*4 = 75 m/s

For finding the distance is stops acceleration, just find d1

d1 = .5(18.75)(4)^2 = 150.

For deacceleration, its just the same equation, except both velocity and acceleration need to be taken into account.

So.
d = .5(A)*t^2 + (vo)*t
v0 = 75
d = .5*(-12.5)*t^2+(75)*6
the velocity is zeo at the stop
v = 0 = At + vo
v = 0 =-12.5*t + 75 m/s
t = 6 seconds

d = .5*(-12.5)*5^2+75*4 = 225

t = 6 seconds, d = 225 meters
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bossyburrito
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9/16/2012 11:48:03 PM
Posted: 4 years ago
"...which is wrong, so it couldn't be 75. I guess I was just figuring it in such a way that it would make sense."
I do that all the time.
#UnbanTheMadman

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Or lose the race to rats
Get caught in ticking traps
And start to dream of somewhere
To relax their restless flight
Somewhere out of a memory of lighted streets on quiet nights..."

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baggins
Posts: 855
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9/17/2012 3:15:09 AM
Posted: 4 years ago
At 9/16/2012 10:07:23 PM, Ren wrote:
Well, coming back to this, I wonder why I started the first second at a stop... but, I guess it's best to always look over your answers to make sure you did them right.

I initially stated:

1 - 0
2 - 25
3 - 50
4 - 75
= 150 + 1,050 = 1,200

So, the answer to a) is 25 m/s, and the answer to b) is 150 m.

Which is completely wrong... it would instead be:

1 - 18.75 +
2 - 37.5 +
3 - 56.25 +
4 - 75 =

187.5 + (75 * 14) = 1237.5

...which is wrong, so it couldn't be 75. I guess I was just figuring it in such a way that it would make sense.

So, I think I'm going to try this on paper, instead, before I go looking for the acceleration transformation that applies to this (you're not left with much information).

...

...well, the direction I ended up going was to reduce it to its smallest increment algebraically and work from there.

What we have here is the initial acceleration applied four times in a stepwise fashion, then that acceleration four times over 14 times repeated... lol... which translates in math terms as:

(x + 2x + 3x + 4x) + 14(4x) = 1,200.
(10x) + 14(4x) = 1,200
10x + 56x = 1,200
66x = 1,200
x = 1,200/66 = 18.1818182

18.1818182 * 4 = 72.7272728

So, in those first four seconds, it was accelerating at 18.1818182 m/s, and it travelled 181.8181819 m by the end of those first four seconds. It thus traveled 72.7272728 m/s thereafter to finally reach 1,200 m at the end of 18 s.

Hilariously, it works out.

It would have been impressive - had you figured out the correct answer in this way!
The Holy Quran 29:19-20

See they not how Allah originates creation, then repeats it: truly that is easy for Allah.

Say: "Travel through the earth and see how Allah did originate creation; so will Allah produce a later creation: for Allah has power over all things.
Ren
Posts: 7,102
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9/17/2012 7:25:25 PM
Posted: 4 years ago
At 9/17/2012 3:15:09 AM, baggins wrote:
At 9/16/2012 10:07:23 PM, Ren wrote:
Well, coming back to this, I wonder why I started the first second at a stop... but, I guess it's best to always look over your answers to make sure you did them right.

I initially stated:

1 - 0
2 - 25
3 - 50
4 - 75
= 150 + 1,050 = 1,200

So, the answer to a) is 25 m/s, and the answer to b) is 150 m.

Which is completely wrong... it would instead be:

1 - 18.75 +
2 - 37.5 +
3 - 56.25 +
4 - 75 =

187.5 + (75 * 14) = 1237.5

...which is wrong, so it couldn't be 75. I guess I was just figuring it in such a way that it would make sense.

So, I think I'm going to try this on paper, instead, before I go looking for the acceleration transformation that applies to this (you're not left with much information).

...

...well, the direction I ended up going was to reduce it to its smallest increment algebraically and work from there.

What we have here is the initial acceleration applied four times in a stepwise fashion, then that acceleration four times over 14 times repeated... lol... which translates in math terms as:

(x + 2x + 3x + 4x) + 14(4x) = 1,200.
(10x) + 14(4x) = 1,200
10x + 56x = 1,200
66x = 1,200
x = 1,200/66 = 18.1818182

18.1818182 * 4 = 72.7272728

So, in those first four seconds, it was accelerating at 18.1818182 m/s, and it travelled 181.8181819 m by the end of those first four seconds. It thus traveled 72.7272728 m/s thereafter to finally reach 1,200 m at the end of 18 s.

Hilariously, it works out.

It would have been impressive - had you figured out the correct answer in this way!

72.7272 * 14 = 1018.18 + 181.8 = 1199.98

Close enough?

Bah.

I think it was close enough. I'll bet it's right, all decimals considered. ;)

So, you can color yourself impressed, or not, over a 0.01 margin of error. I was thrilled nonetheless. ^_^
Ren
Posts: 7,102
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9/17/2012 7:31:44 PM
Posted: 4 years ago
At 9/16/2012 10:35:23 PM, darkkermit wrote:

t = 6 seconds, d = 225 meters

1 - 12.5 + 12.5
2 - 25 + 12.5
3 - 37.5 + 12.5
4 - 50 + 12.5
5 - 62.5 + 12.5
6 - 75

=/=

1,200;

12.5 + 25 + 37.5 + 50 + 62.5 + 75 = 262.5

=/=

225

Wrong answer is wrong?

(btw, I'm pretty sure my answer to that part of the question was right. Care to show how it's not?)
darkkermit
Posts: 11,204
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9/17/2012 10:37:11 PM
Posted: 4 years ago
At 9/17/2012 7:31:44 PM, Ren wrote:
At 9/16/2012 10:35:23 PM, darkkermit wrote:

t = 6 seconds, d = 225 meters

1 - 12.5 + 12.5
2 - 25 + 12.5
3 - 37.5 + 12.5
4 - 50 + 12.5
5 - 62.5 + 12.5
6 - 75

=/=

1,200;

12.5 + 25 + 37.5 + 50 + 62.5 + 75 = 262.5

=/=

225

Wrong answer is wrong?

(btw, I'm pretty sure my answer to that part of the question was right. Care to show how it's not?)

No, because your doing the method wrong.

First of all, for the first part there's no reason why the velocity should equal the distance. The velocity should be zero once it comes to a complete stop. Which it does because the initial velocity is 75 m/s.

Second off, the equation for finding distance is

d = (1/2)A*t^2 + v*t

What your doing is a riemann's sum. This will only get you an approximate, but not the exact.

v = (A)*t + vo

This is the equation your basing the above numbers on. So what your doing is:
d = sum(f(v)*(delta t),0,6)

delta t = 1 second for your calculations, and you've been doing it from 1 to 6 seconds. However there's room for error. Because at 1/2 a second it isn't going at 12.5 seconds. 1.5 seconds isn't going to go at 25 seconds and so forth.

The equation:

d = (1/2)A*t^2 + v*t

is what you use which is actually just the integral of the following:

v = (A)*t + vo
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baggins
Posts: 855
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9/17/2012 10:54:25 PM
Posted: 4 years ago
At 9/16/2012 10:07:23 PM, Ren wrote:
Well, coming back to this, I wonder why I started the first second at a stop... but, I guess it's best to always look over your answers to make sure you did them right.

I initially stated:

1 - 0
2 - 25
3 - 50
4 - 75
= 150 + 1,050 = 1,200

So, the answer to a) is 25 m/s, and the answer to b) is 150 m.

Which is completely wrong... it would instead be:

1 - 18.75 +
2 - 37.5 +
3 - 56.25 +
4 - 75 =

You were correct till here. However you messed up between displacement and acceleration at this point.

Let me explain. Suppose I go from velocity of 50m/s to 80m/s in 3 seconds then:
1. My acceleration is 10 m/s^2 as my velocity increases by 10m/s every second. My speed would be:
0sec -> 50m/s
1sec -> 60m/s
2sec -> 70m/s
3sec -> 80m/s
This is same as what you have done. However you need to notice that 10m/s^2 is acceleration (or the rate at which velocity is increasing).

2. My average speed is (50+80)/2 = 65m/s. Thus my displacement in these 3 seconds is 65*3 = 195m. This is not sum of all velocities in between.

187.5 + (75 * 14) = 1237.5

Adding up velocities does not give you displacement. Use (0+75)/2 as average velocity and 4 sec as time. Turns out this was correct guess.
The Holy Quran 29:19-20

See they not how Allah originates creation, then repeats it: truly that is easy for Allah.

Say: "Travel through the earth and see how Allah did originate creation; so will Allah produce a later creation: for Allah has power over all things.
Ren
Posts: 7,102
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9/18/2012 5:42:42 AM
Posted: 4 years ago
At 9/17/2012 10:37:11 PM, darkkermit wrote:
At 9/17/2012 7:31:44 PM, Ren wrote:
At 9/16/2012 10:35:23 PM, darkkermit wrote:

t = 6 seconds, d = 225 meters

1 - 12.5 + 12.5
2 - 25 + 12.5
3 - 37.5 + 12.5
4 - 50 + 12.5
5 - 62.5 + 12.5
6 - 75

=/=

1,200;

12.5 + 25 + 37.5 + 50 + 62.5 + 75 = 262.5

=/=

225

Wrong answer is wrong?

(btw, I'm pretty sure my answer to that part of the question was right. Care to show how it's not?)

No, because your doing the method wrong.

First of all, for the first part there's no reason why the velocity should equal the distance. The velocity should be zero once it comes to a complete stop. Which it does because the initial velocity is 75 m/s.

Second off, the equation for finding distance is

d = (1/2)A*t^2 + v*t

What your doing is a riemann's sum. This will only get you an approximate, but not the exact.

v = (A)*t + vo

This is the equation your basing the above numbers on. So what your doing is:
d = sum(f(v)*(delta t),0,6)

delta t = 1 second for your calculations, and you've been doing it from 1 to 6 seconds. However there's room for error. Because at 1/2 a second it isn't going at 12.5 seconds. 1.5 seconds isn't going to go at 25 seconds and so forth.

The equation:

d = (1/2)A*t^2 + v*t

is what you use which is actually just the integral of the following:

v = (A)*t + vo

Thanks! ^_^
Ren
Posts: 7,102
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9/18/2012 5:47:14 AM
Posted: 4 years ago
At 9/17/2012 10:54:25 PM, baggins wrote:
At 9/16/2012 10:07:23 PM, Ren wrote:
Well, coming back to this, I wonder why I started the first second at a stop... but, I guess it's best to always look over your answers to make sure you did them right.

I initially stated:

1 - 0
2 - 25
3 - 50
4 - 75
= 150 + 1,050 = 1,200

So, the answer to a) is 25 m/s, and the answer to b) is 150 m.

Which is completely wrong... it would instead be:

1 - 18.75 +
2 - 37.5 +
3 - 56.25 +
4 - 75 =

You were correct till here. However you messed up between displacement and acceleration at this point.

Let me explain. Suppose I go from velocity of 50m/s to 80m/s in 3 seconds then:
1. My acceleration is 10 m/s^2 as my velocity increases by 10m/s every second. My speed would be:
0sec -> 50m/s
1sec -> 60m/s
2sec -> 70m/s
3sec -> 80m/s
This is same as what you have done. However you need to notice that 10m/s^2 is acceleration (or the rate at which velocity is increasing).

2. My average speed is (50+80)/2 = 65m/s. Thus my displacement in these 3 seconds is 65*3 = 195m. This is not sum of all velocities in between.

187.5 + (75 * 14) = 1237.5

Adding up velocities does not give you displacement. Use (0+75)/2 as average velocity and 4 sec as time. Turns out this was correct guess.

Oh yeah, you're absolutely right! I can't believe I made that mistake. :P

Thanks! ^_^