Really easy thing but i'm getting something wrong

  • Thread starter tgoot84
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  • #1
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In a grocery store, you push a 14.5 kg shopping cart with a force of 11.0 N. If the cart starts at rest, how far does it move in 3.00 s?

i've got that acceleration = force/mass

so the acceleration i got was .7586
but for some reason i keep getting the answer to the problem wrong

theanswer is in meters

_____m
 

Answers and Replies

  • #2
156
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It would be good to show us what you are getting so we may be able to help you better. Post your work with the incorrect answer you get.

Try this kinematic equation:
[tex]x_{f}=x_{i} + v_{i}(t) + \frac{1}{2}a(t)^2[/tex]
 
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  • #3
HallsofIvy
Science Advisor
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Is this a college or secondary school question?

Because you posted it in both areas!
 
  • #4
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Paul could you or Halls or somebody work it out and talk tgoot84 (and me) through it? I thought I could help but it turns out I can't remember how to work these, either.

Okay, tgoot84 got the acceleration by using a=f/m. It's .7586. Now what?

The formula you gave, we change the acceleration into an average velocity by subtracting zero from .7586, and then .7586 x 3 seconds / 2 ? And then use that rate to multiply by the seconds squared (9)?

Giving how far it went in 3 seconds?
 
  • #5
156
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Holly,
If you use the kinematic equation I posted, you do not need average velocity at all. In fact, you don't need velocity period.

In the above equation, start with the initial position xi. Decide where it starts from , so let that be zero.
You also know the initial velocity, vi, it is zero. This is from the question stating that the cart is at rest. So the whole middle term, vi(t) drops out too.

That leaves [tex]x_{f}=\frac{1}{2}a(t)^2[/tex]

Now you plug in and calculate. You know the acceleration, it was found with a=F/m. You know the time, 3.00s from the problem. Units of seconds will cancel out, and your final answer will be in meters.

Work it out and post back.
 
  • #6
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Okay, I changed the numbers because I like to work with tens and things.

I made my cart weigh 10kg. I made the F be 100N. a=F/m so a=10m/s/s.

Your equation, I write it as d=1/2at^2. That "x" is confusing!

d=1/2a(t^2). d=5(t^2). d=5 times 3^2. d=5 times 9. So, it went 45m. My cart, that is.

Thank you for the help, don't know why I have to constantly relearn things. But COULD I have gotten the answer with the average velocity and the d=rt formula? d=r, the r is 10m/s/s times the 3 seconds, minus the zero meters it went because it wasn't moving, divided by 2 equals 15, now I multi it by 3, and it says 45. Well, that looks the same.
 
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  • #7
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Huh!!!???

How do you change the force from 11 to 100? How is that easier than working with the numbers given. Your acceleration ends up being 10m/s^2!? This is way off from the .7586 that is found with the actual numbers given.

I use x because it is easy to think of it as the cart moving horizontally along the x axis.

The formula you end up with is x = .5* acceleration * time^2

x = .5 * .7586 * 3^2
x = .5 * .7586 * 9
x = .3793 * 9
x = 3.41

If you carried the units through with the values, you will see that it ends up as (m/s^2)*s^2. You can see here that the s^2 terms cancel out, and meter is left.

Let me know that you got this.
 
  • #8
184
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Geewillikers! What does it matter what the numbers are? I just like to learn things with easy numbers, so I can see what I'm doing!

And I did too get the right answer with the d=rt formula! I used my nice 10 and 100 numbers FOR EASY FIGURING.
First I found r! rate = 30m = 0 meters divided by 2 to get average velocity. That's 15m/s. Then, I just put it in the d=rt formula.
d=15 x 3 seconds. That's 45 m/s. That's right!!!

I don't like your way, Paul, because it confuses me, the part where the 1/2at^2 is! I do that Please-Excuse-My-Dear-Aunt-Polly to get the order of operations, so I square whatever t is first. But then, what about the a and the 1/2? I guess I have to divide the a by 2 first, then multiply that by whatever the time squared was, or do I do all the multiplying first, and then divide the whole thing by 2? It's just confusing!
 
  • #9
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Using easy numbers to learn a concept or approach is fine, but your answers are not correct for this problem.

How did you find the rate? The final speed of the cart, at 3 seconds, is not 30m. Not only is this value wrong, but so are the units. Where did this come from? A speed (velocity) will have units of meters/seconds (distance divided by time).
The final velocity of this cart at 3 seconds is 2.28m/s, not 30m.

Your method of finding average speed sounds right. The average speed here is 2.28 / 2 = 1.14m/s

If you want to use average speed to find the distance it traveled, then 1.14m/s * 3s = 3.42m/s

I am sorry if my previous explanation seems confusing, but it is the most direct, and simple, way to solve this problem. By finding the average speed you are adding calculations. You do not have to find the speed.

If it seems like I am nit picking, I am not. It is important to pay attention to the details. Things like the units are important.
 
  • #10
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We were looking for distance, weren't we, originally, in tgoot84's problem?

The cart was 14.5kg, the force was 11N, and it moved, initially from rest, for 3 seconds. How far did it go by the end of the 3 seconds, right?

d=1/2at^2. Find acc first: a=f/m. So, a=11N/14.5kg, = 0.7586
Substitute that back in:
d=1/2(.7586)(3^2)
d=0.3793(9)
d=3.4 m

or,
d=rt
find r first: find acc, then turn it into avg velocity.
a=F/m, so a=11n/14.5kg, it's 0.7586 as it was above.
It moved for 3 seconds, so times 3 that gives 2.2758. Divided by 2, that's 1.137
d=1.137m/s(3 seconds)
d=3.41m

When I used 10kg and 100n, the answer was 45 meters.
I am shaky on the units, but I still did too get the answer!!! I hope tgoot84 understands it now, too.
 
  • #11
156
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I don't know what is going on here, but you have just made a complete turn around...

The cart was 14.5kg, the force was 11N, and it moved, initially from rest, for 3 seconds. How far did it go by the end of the 3 seconds, right?

d=1/2at^2. Find acc first: a=f/m. So, a=11N/14.5kg, = 0.7586
Substitute that back in:
d=1/2(.7586)(3^2)
d=0.3793(9)
d=3.4 m

You went from having no clue, to mimicking my method perfectly?

One thing still stands. You posted that the cart traveled 45 meters in 3 seconds. How does 45 meters equate to 3.41 meters? Do you think these values are equivalent?
 
  • #12
Going from not having a clue to a proper method is the whole point of learning, isn't it?

cookiemonster
 
  • #13
184
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Okay, ONE MORE TIME!!!
I did TWO problems, ONE, tgoot84's problem with weird numbers. THEN, another problem with EASY NUMBERS, all nice 10's and 100's, that I MADE UP.

For MY problem, the distance traveled was 45m. For tgoot84's problem it was 3+meters...

I really hate how suspicious people are of me.

I learn a certain way. I must SEE it done, SEE it and look at it and look at SEVERAL of it before I can do it. Once I can SEE it with more than one set of numbers, I can usually DO it. I can't learn to do it unless it is worked out; in fact, there must be several examples worked out. Surely I'm not the only person who learns this way? BUT, Tomorrow, or maybe later today, I will again have to SEE AT LEAST TWO EXAMPLES OF PROBLEMS WITH REAL NUMBERS, NOT JUST THE FORMULA LETTERS/SYMBOLS, in order to reproduce the method again myself. I am always so glad when two people answer my questions, because then I get more to see.

I was so happy to find this forum! And then, very gloomy-minded people regard me as insincere. I met a woman who made Ukrainian eggs: Pysanki. Okay, she agreed to teach me. I watched her make one egg. I watched her make two eggs. Then I picked up a kistky and I made an egg, better than her eggs, okay? She was MAD, yelled that I *had* to have already known how to make the eggs, COULD NOT have learned it that quickly. Well, BULLCORN! Same in a French class, the T.A. screamed I had to have taken French before. It's very disheartening.
 
  • #14
Don't worry about it, holly. There's no need for you to defend yourself.

cookiemonster
 
  • #15
156
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Many people learn in a similar way. I have to drill practice problems to make things stick.

The problem here is that I had no idea what you were doing until now. You acted as if I was sitting there with you while you worked through two seperate problems.

It would have helped greatly if you had explained that to me earlier instead of letting me think you were pulling numbers out of thin air, and then proclaiming the ANSWER to be something from your private problem!? How was I to know?

I kept coming back with responses as an earnest attempt to make sure you got it. Now you get down on me for it?

I am also happy to have found this forum, and I am trying to give back what I can.

Cookiemonster - Of course that is the whole point of learning. I just wish that I had a better understanding of what she was doing. I hope she masters the question, and all problems she tackles in the future. Please read the entire forum and tell me if you followed what she was doing.

And you are correct. She does not need to defend herself. Holly will tell you herself that I have offered nothing but encouragement. I only ask that we communicate a little better.
 
  • #16
Personally, I've been following this little discussion. I thought it was pretty clear what holly was doing, but I've seen the same thing done numerous times and used to have a tendency to do it myself. It is most certainly true that her words could have been interpreted in the way that you did, paul, so there is no criticism of you. I'm certain your contributions are appreciated.

I suppose the only thing you could have done is tried to consider some alternative possibilities when what she was doing when it didn't make sense to you.

As for "reading the entire forum," it's quite a challenge, but I'm working at it. ;) I don't think I could ever make it through all the theory discussions and politics discussions, but a lot of them are really interesting.

cookiemonster
 
  • #17
156
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As for reading the entire forum...

Touché...
 
  • #18
Just a joke...

I've been known to say some goofy things myself.

cookiemonster
 
  • #19
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The last thing I want is to offend people who have helped me. Paul has helped me very, very much. But I don't like it when it's intimated that I'm some sort of physics "ringer," someone insincere. I take so much guff amongst the kiddies at the college already...guess I am almost over-sensitive to it. I should have been plainer what I was trying to do, but truly, sometimes I can't be plain; I hardly know what I'm doing.

On to worse problems than my defective personality:

I can't understand a set of problems in the workbook. It's three drawings of a speeding car. The first car says v=30km p h and under that car, it says, K.E.= 10^6 Joules. The next car says v=60kph, and it's blank underneath. The third car says v=90kph, and it also has a fill-in-the-blank. I tried to do the K.E.=1/2mv^2, but got giant, giant numbers, like 10 million. What am I doing wrong? I am still on chapter 3 and the class is on chapter 7 now.
 
  • #20
jamesrc
Science Advisor
Gold Member
476
1
Hello,

There are two ways to do this problem you have posted (well, 2 ways that jump out at me; there's almost always a number of ways to solve any given problem). I'm assuming that these are all the same car; that is they all have the same mass = m. The energies are in the millions of Joules, so you may be doing this correctly after all.

Anyway, the first way is to use [tex] {\rm KE} = \frac{mv^2}{2} [/tex] as you said. For the first car, you are given the energy and the velocity, so you want to find the mass, m. When you have the mass, you can plug that back into the next 2 cars (along with their velocities) to find the energies there. Before you do this, you must notice the units on the velocity. In order to use the formula for energy and get it in units of Joules, the velocity has to be in meters per second. So as an example:

[tex] 30 \frac{\rm km}{\rm h}\cdot\frac{1000\rm m}{\rm km}\cdot\frac{\rm h}{3600\rm s} = 8.33\frac{\rm m}{\rm s}[/tex]

(notice how the units cancel out.)

The second way may not be as immediately obvious to you. Let's label some variables to start off with:

K1: Kinetic energy of car 1
K2: Kinetic energy of car 2
K3: Kinetic energy of car 3
v1: speed of car 1
v2: speed of car 2
v3: speed of car 3

We know that [tex] K_i = \frac{mv_i^2}{2} [/tex], where i can be 1, 2, or 3.
Look at K2:
[tex] K_2 = \frac{mv_2^2}{2} [/tex]
which is the same as:
[tex] K_2 = \frac{mv_1^2}{2}\left(\frac{v_2}{v_1}\right)^2 [/tex]
but since
[tex] K_1 = \frac{mv_1^2}{2} [/tex]
that means:
[tex] K_2 = K_1\left(\frac{v_2}{v_1}\right)^2 [/tex]

All of which is just explicitly showing that if you double the velocity, you quadruple the energy. If you triple the velocity, you multiply the energy by nine.
 
  • #21
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Jamesrc, thank you for the help. I did not realize about making the units match...
So, by that second method you showed, I can just increase the first answer (the given answer of 10^6) by 4 times & 9 times, as in 10^24 and 10^54?

Thanks,
Holly
 
  • #22
Be careful. 4 times 10^6 is 4*10^6, not (10^6)^4.

cookiemonster
 

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