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Acceleration and displacement

  • #1

Homework Statement


A car accelerates uniformly from rest along a straight road and experiences negligible frictional forces. The car travels 15m in the second second of its journey. How far does it travel in the fifth second? (OPTIONS: 15m, 35m, 45m, 75m)

Homework Equations


3. The Attempt at a Solution [/B]

This seems really sophomore but I'm not arriving at any of those options. What I did was to consider that 15m was covered in two seconds (though I'm not completely sure if that's what is said, or if it means to say that 15m was covered in that specific second). So 15 / 22 = 3.75 ms-2. Multiplying this by 52 gives 93.75m.
Plan B, assume 15m was covered in the specific second. Using a = (v - u) / t, v = 15ms-1, u = 0 and t = 2, giving 7.5ms-2.
7.5 * 52 = 187.5s.
?
 

Answers and Replies

  • #2
gneill
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The second second of its journey covers the time period t = 1 s to t = 2 s. The fifth second covers the interval t = 4 s to t = 5 s.

Don't forget the velocity that the car has when it begins each interval.
 
  • #3
The second second of its journey covers the time period t = 1 s to t = 2 s. The fifth second covers the interval t = 4 s to t = 5 s.

Don't forget the velocity that the car has when it begins each interval.
I don't follow. What's the error in my working? The units work out.
 
  • #4
gneill
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I don't follow. What's the error in my working? The units work out.
The car did not cover 15 m in two seconds. It covered 15 m during the second second. That is, from time t = 1s to time t = 2s.
 
  • #5
The car did not cover 15 m in two seconds. It covered 15 m during the second second. That is, from time t = 1s to time t = 2s.
Right, so u =/= 0. I'm not sure how to find u then.
 
  • #6
gneill
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Work symbolically. Assuming a constant acceleration a, what would be the velocity at the end of the first second?
 
  • #7
Work symbolically. Assuming a constant acceleration a, what would be the velocity at the end of the first second?
a = v1 - 0 / 1
a = v1?
----
a = 15 - v1
a = v1
2v1 = 15
v1 = a = 7.5
No option for 7.5m. 7.5 * 5 = 37.5m.
 
  • #8
gneill
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a = v - 0 / 1
a = v?
Working without units that is fine. A marker may be picky about dropping units, so you might want to declare a variable to represent a time interval of one second. Say, Δt = 1 s. Then you could write:

##v1 = a Δt##

So, v1 will be the speed of the car as it begins its second second of the journey. What equation of motion can you write to find the distance traveled over that one second interval?
 
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  • #9
Working without units that is fine. A marker may be picky about dropping units, so you might want to declare a variable to represent a time interval of one second. Say, Δt = 1 s. Then you could write:

##v1 = a Δt##

So, v1 will be the speed of the car as it begins its second second of the journey. What equation of motion can you write to find the distance traveled over that one second interval?
Hmm. How was what I wrote wrong in practice?
s = 0.5 * a * 12.
I don't understand. I'd need one more value.
 
  • #10
gneill
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Hmm. How was what I wrote wrong in practice?
s = 0.5 * a * 12.
I don't understand. I'd need one more value.
That's not the whole of the basic equation of motion. What happened to the initial velocity?
 
  • #11
That's not the whole of the basic equation of motion. What happened to the initial velocity?
It's 0, because I'm looking at the first second as you suggested, and the car starts at rest.
If you were to look at the second second, you'd have s = v1 + 0.5v1 = 1.5v1, assuming that v1 = a.
EDIT: Ah, s = 15. So v1 = 10m/s?
Just a note, I didn't think the question would be this long-winded. It's a 1 mark and this seems beyond what we've been taught.
How was my value for 7.5ms-2 wrong? Or was it right?
 
  • #12
gneill
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It's 0, because I'm looking at the first second as you suggested, and the car starts at rest.
If you were to look at the second second, you'd have s = v1 + 0.5v1 = 1.5v1, assuming that v1 = a.
Just a note, I didn't think the question would be this long-winded. It's a 1 mark and this seems beyond what we've been taught.
It will seem simple once you've had the "aha!" moment.

Perhaps I wasn't clear in post #8. What I meant was, what is the equation of motion for the second one second interval (the "second second" so to speak), given that you now know the velocity going into that second is ##v1 = a \Delta t## ?
 
  • #13
It will seem simple once you've had the "aha!" moment.

Perhaps I wasn't clear in post #8. What I meant was, what is the equation of motion for the second one second interval (the "second second" so to speak), given that you now know the velocity going into that second is ##v1 = a \Delta t## ?
Maybe, but now I have to understand why all my other working was wrong. :P Why was the 7.5 wrong?

I edited my post just before yours, is that progress?
 
  • #14
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If a is the acceleration and t is the time, what is the equation for the total distance s up to time t (assuming it starts from rest)?
In terms of a, what is the total distance covered in 2 sec?
In terms of a, what is the total distance covered in 1 sec?
In terms of a, what is the total distance covered in 5 sec?
In terms of a, what is the total distance covered in 4 sec?
 
  • #15
If a is the acceleration and t is the time, what is the equation for the total distance s up to time t (assuming it starts from rest)?
In terms of a, what is the total distance covered in 2 sec?
In terms of a, what is the total distance covered in 1 sec?
In terms of a, what is the total distance covered in 5 sec?
In terms of a, what is the total distance covered in 4 sec?
s = 0.5 * a * t2? I dunno, lol

haruspex, would greatly appreciate your help if you're reading.
 
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  • #16
gneill
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Maybe, but now I have to understand why all my other working was wrong. :P Why was the 7.5 wrong?

I edited my post just before yours, is that progress?
Without units or comments explaining what the steps are, it is difficult for me to follow what you are trying to do there. When you replace a velocity with "a" because it's really "a ⋅ 1s" without a note stating that, things get muddy. That's why I suggested using the variable ##\Delta t## to represent a time interval of 1 second. Either hang on to the units or use variables to carry them along.

Let's try a slightly different approach that might avoid some of the syntax issues.

Since the vehicle is starting from rest and the acceleration is constant, the distance traveled versus time is simply:

##d(t) = \frac{1}{2} a t^2##

Now you are given a distance traveled during a specific time interval, namely the second second of the journey. So use the above equation to find the expressions for the distances at either end of that interval. The difference between them should be your given value of distance for that interval. The only variable should be the acceleration which you can solve for.
 
  • #17
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s = 0.5 * a * t? I dunno, lol
How can you possibly expect to be able to solve a problem like this if you can't even correctly state this basic formula?? lol, lol, lol.
 
  • #18
How can you possibly expect to be able to solve a problem like this if you can't even correctly state this basic formula?? lol, lol, lol.
Throwing a cheap blow over a technicality? It was bad enough that you were useless.

Without units or comments explaining what the steps are, it is difficult for me to follow what you are trying to do there. When you replace a velocity with "a" because it's really "a ⋅ 1s" without a note stating that, things get muddy. That's why I suggested using the variable ##\Delta t## to represent a time interval of 1 second. Either hang on to the units or use variables to carry them along.

Let's try a slightly different approach that might avoid some of the syntax issues.

Since the vehicle is starting from rest and the acceleration is constant, the distance traveled versus time is simply:

##d(t) = \frac{1}{2} a t^2##

Now you are given a distance traveled during a specific time interval, namely the second second of the journey. So use the above equation to find the expressions for the distances at either end of that interval. The difference between them should be your given value of distance for that interval. The only variable should be the acceleration which you can solve for.
The only variable would be acceleration, but I don't know the initial velocity between t = 1 and t = 2?
15 = u + 0.5 * a (for 2nd second, t = 1 so neglected)
s = 0.5 * a (for 1st second)
I need to understand why I was wrong for both the 7.5ms-2 = a value and the 10ms-1 = v1 final velocity after 1st second for this to click. Here's what I was doing in bold:
a = v1 - 0 / 1 (for 1st second)
a = v1? (u = 0 and t = 1 so a works out to be the final velocity after the 1st second)
----
a = 15 - v1 (for 2nd second, 15m/s - initial velocity is the acceleration)
a = v1
2v1 = 15 (substitution)
v1 = a = 7.5
No option for 7.5m. 7.5 * 5 = 37.5m.
 
  • #19
gneill
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The only variable would be acceleration, but I don't know the initial velocity between t = 1 and t = 2?
You won't need it with this method. You can write explicit expressions for the total distance traveled up to any given time. Write the expression for t = 1 second and the expression for t = 2 seconds. The difference is the distance covered during the interval from t = 1 second to t = 2 seconds.
 
  • #20
You won't need it with this method. You can write explicit expressions for the total distance traveled up to any given time. Write the expression for t = 1 second and the expression for t = 2 seconds. The difference is the distance covered during the interval from t = 1 second to t = 2 seconds.
The distance s travelled after 1 second is s = 0.5 * a
The distance s travelled after 2 seconds is s = 2u + 0.5 * a * 22
2u + 0.5 * a * 22 - 0.5 * a = 2u + 1.5a = 15 (second distance - first distance)
Is that what you meant?
This is really too much for a 1 mark.
 
  • #21
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Throwing a cheap blow over a technicality? It was bad enough that you were useless.
I apologize. It's just that you seemed so cavalier about it.

I think if you reconsider answering my questions in post #14, you will find that it was not as useless as you originally thought, and that it will actually help you solve your problem. It is very close to what gneill was trying to help you do.

Again, sorry for not having more patience.

Chet
 
  • #22
gneill
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The distance s travelled after 1 second is s = 0.5 * a
The distance s travelled after 2 seconds is s = 2u + 0.5 * a * 22
2u + 0.5 * a * 22 - 0.5 * a = 2u + 1.5a = 15 (second distance - first distance)
Is that what you meant?
Almost, but there's no initial velocity involved. The expression giving total distance does not need a "u", since the car starts from rest. u = 0 always in this case. So write the second expression without the u. Then take the difference and find a.
This is really too much for a 1 mark.
I think you're making it more difficult than it needs to be. If you understand that the expression ##d(t) = \frac{1}{2}a t^2## applies for all time when the car starts from rest, then you wouldn't introduce "u" along the way to complicate things. You only need to find the difference between two distances, one at each end of the the intervals in question.
 
  • #23
I apologize. It's just that you seemed so cavalier about it.

I think if you reconsider answering my questions in post #14, you will find that it was not as useless as you originally thought, and that it will actually help you solve your problem. It is very close to what gneill was trying to help you do.

Again, sorry for not having more patience.

Chet
I apologise in turn. I was cavalier about it, because I thought that this was getting seriously overblown for a 1 mark. But I've just learnt that this question has been taken from a practice guide, not an exam paper. It would probably pass for more in an exam.

Anyway, I thought about this some more. (Note to self: it helps to have a paper to scribble on rather than wrestling everything out in the mind.)
The distance travelled in the first 2 seconds = 0.5 * a * 22 = 2a
The distance travelled in the first 1 second = 0.5 * a * 12 = 0.5a
The first distance minus the second distance should give me 15m.
2a - 0.5a = 1.5a = 15m. a = 10 ms-2.
Therefore the distance travelled in each of the 5 seconds should go something like:
S1 = 5m
S2 = 15m
S3 = 25m
S4 = 35m
S5 = 45m.
The sum of all of these is 125m, which is consistent with s = 0.5 * 10 * 52 = 125.

So I'd say the answer is 45m. The mark scheme disagrees with me and says it's 75m. Yet I took this to both of my physics teachers, separately, and they agreed with my line of reasoning and said the mark scheme could well be wrong -- again, this was just a practice set taken from online.

Even if I'm correct, I have some doubts:
1) If the acceleration is in fact 10, why wouldn't the distance covered in the first second from rest be 10? Acceleration is the change in velocity per second and the car starts at rest.
2) Wouldn't the distance covered in the first two seconds = 3a? a would be covered in second 1, a + a would be covered in second 2. Maybe I'm making the same fallacy in this and the above question.
3) If we consider the first two seconds, v = 15m/s, u = 0 and t = 2s. (15 - 0) / 2 = 7.5ms-2. I'd guess the error is in taking 15m/s for v, but I'm not sure why.

Thanks to both of you for getting me this far, and thanks to anyone who has the patience to respond to this.
 
  • #24
gneill
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Acceleration represents a change in velocity, not a change in distance. In the first second the velocity goes from 0 m/s to 10 m/s, but it is not 10 m/s over the whole first second. In fact its average velocity is 5 m/s over that first second.
 
  • #25
Acceleration represents a change in velocity, not a change in distance. In the first second the velocity goes from 0 m/s to 10 m/s, but it is not 10 m/s over the whole first second. In fact its average velocity is 5 m/s over that first second.
Helpful, thanks. This answers 1) and 2), still not sure about 3) because the final velocity would still be 15m/s. That wasn't questioned earlier.

Also, I take it you're happy with my answer and working?
 

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