How to determine the car's deceleration at a point along a curve?

In summary, the conversation was about finding the angle of a circular sector in a physics problem. The speaker had difficulty understanding how the angle of the sector could be determined to be 30 degrees, and another person explained that it was due to the tangent and normal lines being perpendicular to each other. The theorem of angles with perpendicular sides being equal was also mentioned as a useful tool in physics problems. The conversation ended with the speaker gaining a better understanding of how to determine angles in physics problems.
  • #1
simphys
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Homework Statement
When the car reaches point A it has a speed of ##25m/s##. If the brakes are applied, its speed is reduced by ##a_t = -\frac 14 \sqrt{t} \frac {m}{s^2}##. Determine the magnitude of acceleration of the car just before it reaches point ##C##.
Relevant Equations
##s = R*\theta##
So I get the exercise and all and have just solved it. But .. I kind of very very intuitively determined ##\theta## to also be the angle for the circular sector.
The problem here is that my geometry bag is very weak, I didn't have any geometry in HS, will fix that sooner or later but anyway. so... My question is.. how do I 'actually' derive that the angle of the circular sector equals ##\theta = 30## degrees in such situations?

Thanks in advance.
My thought process was, okay so theta keeps increasing as we move along the circular segment which probably implies that the angle of the sector ##= \theta##
1658139069204.png

note: the circular path is from B --> C. It might not look as it is due to the placement of the center of curvature, but it is.
 

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  • #2
simphys said:
So I get the exercise and all
Good. We can't: the image is severly blurred and illegible.
What is the complete problem statement ?
 
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  • #3
BvU said:
Good. We can't: the image is severly blurred and illegible.
What is the complete problem statement ?
I have written in the problem statement section?? what do you mean
The picture is just for the drawing of the situation.
 
  • #4
BvU said:
Good. We can't: the image is severly blurred and illegible.
What is the complete problem statement ?
it's reaaaal bad.. let me take another one.
 
  • #5
BvU said:
Good. We can't: the image is severly blurred and illegible.
What is the complete problem statement ?
edited, thank you.
 
  • #6
simphys said:
that the angle of the sector =θ
That's correct. The heading of the car changes bij 30 degrees.

[edit] The ##\rho=250\ \text m## appears to be not to scale (the distance to ##B## is much more than to ##C## ). A bit misleading.

##\ ##
 
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  • #7
BvU said:
That's correct. The heading of the car changes bij 30 degrees.

##\ ##
yes but I don't get why that implies that the angle of THE CIRCULAR SEGMENT also equals ##30## degrees.
 
  • #8
BvU said:
That's correct. The heading of the car changes bij 30 degrees.

[edit] The ##\rho=250 \text m## appears to be not to scale (the distance to ##B## is much more than to ##C## ). A bit misleading.

##\ ##
yep I noted it under the picture already :), thank you for mentioning.
 
  • #9
simphys said:
yes but I don't get why that implies that the angle of THE CIRCULAR SEGMENT also equals ##30## degrees.
The tangent to the circle is always at right angles to the radius. Increase one and the other will increase by the same amount/. Is that your problem?
 
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  • #10
sophiecentaur said:
The tangent to the circle is always at right angles to the radius. Increase one and the other will increase by the same amount/. Is that your problem?
Oh, so basically.. the angle b/n two tangents is equivalent to the angle between normals of the tangents? Is that what you meant? And then from this we know that it's going to be the angle of the circular segment.
 
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  • #11
simphys said:
Oh, so basically.. the angle b/n two tangents is equivalent to the angle between normals of the tangents? Is that what you meant? And then from this we know that it's going to be the angle of the circular segment.
Yes. It's simple geometry. The two lines are fixed with 90° between them. Increase one by x and the other will increase by x. Imagine a set square rotating around a pivot on one arm. Every part of it will be rotating by the same angle
 
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  • #12
sophiecentaur said:
Yes. It's simple geometry. The two lines are fixed with 90° between them. Increase one by x and the other will increase by x. Imagine a set square rotating around a pivot on one arm. Every part of it will be rotating by the same angle
unfortunately not so simple for me 😬
but yeah I kind of get it now I think, thanks a lot.
 
  • #13
There is a theorem in geometry: angles with perpendicular sides are equal. This means that if the sides of one angle are each perpendicular to one of the sides of the other angle, the two angles are equal. It is quite useful in physics problems.
 
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  • #14
nasu said:
There is a theorem in geometry: angles with perpendicular sides are equal. This means that if the sides of one angle are each perpendicular to one of the sides of the other angle, the two angles are equal. It is quite useful in physics problems.
Thank you, I just realize that I do the exact same thing with the mg-force, I always translate the perpendicular axis on the other axis that has slope i.e. if the x-axis is rotated to have a slope-angle --> the y-axis will also rotate in the same amount in the same direction. From this I then know that at which side the angle'll be for the mg-force.
 
  • #15
nasu said:
There is a theorem in geometry: angles with perpendicular sides are equal. This means that if the sides of one angle are each perpendicular to one of the sides of the other angle, the two angles are equal. It is quite useful in physics problems.
Oh dear. Whatever is an angle with a perpendicular side? I'm sure you are referring to something that I learned over sixty years ago but I can't think what it is.
 
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  • #16
nasu said:
There is a theorem in geometry: angles with perpendicular sides are equal. This means that if the sides of one angle are each perpendicular to one of the sides of the other angle, the two angles are equal. It is quite useful in physics problems.
@nasu wait.. you are asuming another, is this one by chance? because these are really the only two that I know to be quite honest just as they're needed a lot indeed. let me show ya.
 
  • #17
1658174830322.png

so from that theorem we get ##\phi## == ##\theta##, correct?
 
  • #18
Look at the left hand side of the image linked below. This is what I was talking about.
https://etc.usf.edu/clipart/48200/48228/48228_angperpsides.htm
The angles 7 and and 1 in the fiure on the left hand side are like this and they are equal. Each side of the angle 7 is perpendicular to one od the two sides of the angle 1. The sides are perpendicluar and not the angles.
 
  • #20
sophiecentaur said:
Oh dear. Whatever is an angle with a perpendicular side? I'm sure you are referring to something that I learned over sixty years ago but I can't think what it is.
It is long ago, true. I learned about it in grade 7 or 8. But I used it many times in physics problems so it stayed fresh. And I tell about it to the first years students who nowadays have almost no geometry in high school.
 
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  • #21
sophiecentaur said:
Oh dear. Whatever is an angle with a perpendicular side? I'm sure you are referring to something that I learned over sixty years ago but I can't think what it is.
hahahaha
 
  • #22
nasu said:
It is long ago, true. I learned about it in grade 7 or 8. But I used it many times in physics problems so it stayed fresh. And I tell about it to the first years students who nowadays have almost no geometry in high school.
well @nasu to be honest, my problem was that I didn't have math for approx. 1.5 years for the last 3 years(including geometry) and next to that I didn't have any science and also my Math was only 2 hrs a week.
So it's basically all self-teaching, but haven't reached geometry yet. Skipped it due to time constraints the past year.
 
  • #23
simphys said:
View attachment 304351
so from that theorem we get ##\phi## == ##\theta##, correct?
That can only be correct if the top and bottom lines are parallel. For any other angle they are not equal. I think you are trying to apply properties of similar triangles but is something missing in the diagram (on the top left edge of the green bit)?
nasu said:
And I tell about it to the first years students who nowadays have almost no geometry in high school.
You seem to be telling them something that's wrong. Euclidean Geometry is totally self consistent and everything fits the rules. What actual rule are you trying to apply here?
 
  • #24
simphys said:
hahahaha
You are laughing but what I learned was the real deal. You seem to be making some of your stuff up and that is disastrous in pure geometry. If you can't prove it from first principles it's wrong. Can you prove what you are claiming about what's on the green paper?
 
  • #25
sophiecentaur said:
You are laughing but what I learned was the real deal. You seem to be making some of your stuff up and that is disastrous in pure geometry. If you can't prove it from first principles it's wrong. Can you prove what you are claiming about what's on the green paper?
I am sorry.. but have you read what I have said.. I have NEVER studied geometry (yet). so I do NOT know those first principle derivations. I know that it is important in physics, but I'll come that later on as I have referred to in post ##22

So inevitably as I haven't studied the real deal yet, I will make assumption that are just wrong ain't that right?
 
  • #26
sophiecentaur said:
You seem to be telling them something that's wrong. Euclidean Geometry is totally self consistent and everything fits the rules. What actual rule are you trying to apply here?
Why is it wrong? It is part of Euclidian geometry.
https://etc.usf.edu/clipart/70000/70087/70087_anglesum.htm
You mean that they can also be supplementary? This does not make it wrong. We are not talking proof of a geometrical theorem but just the use of a tool to figure angles in physics problems.
 
  • #27
nasu said:
Why is it wrong?
It's wrong because your green diagram is not the same as the diagram in the link. Only one of your pairs of sides are perpendicular whereas both pairs of sides are perpendicular in the link diagram. A'C' and AC are perpendicular but also AB and A'B' are perpendicular. Your diagram only has one pair of such lines. You could 'wave' that top line and get any value for φ and wave the bottom line to get any value of θ. If you actually make φ equal to θ then you get your perpendiculars and the theorem is satisfied.
 
  • #28
sophiecentaur said:
It's wrong because your green diagram is not the same as the diagram in the link. Only one of your pairs of sides are perpendicular whereas both pairs of sides are perpendicular in the link diagram. A'C' and AC are perpendicular but also AB and A'B' are perpendicular. Your diagram only has one pair of such lines. You could 'wave' that top line and get any value for φ and wave the bottom line to get any value of θ. If you actually make φ equal to θ then you get your perpendiculars and the theorem is satisfied.
I don't have any green diagram. You may have mixed the posts. I showed the poster of the green diagram the actual theorem. As I showed you too
 
  • #29
nasu said:
I don't have any green diagram. You may have mixed the posts. I showed the poster of the green diagram the actual theorem. As I showed you too
Sorry but my issue is with the ‘green’ diagram, posted higher up and which the poster is trying to relate ‘your’ theorem, wrongly. That link you posted is useful and shows he’s not correct. Ah yes - the poster was @simphys
 
  • #30
sophiecentaur said:
Oh dear. Whatever is an angle with a perpendicular side? I'm sure you are referring to something that I learned over sixty years ago but I can't think what it is.
this is when two lines meet, they form right angles.
 
  • #31
JesseProbst said:
this is when two lines meet, they form right angles.
Unless you state the whole thing, it’s meaningless. There are two pairs of lines involved. Read that helpful link. It states the actual theorem.
 
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  • #32
Err, ok, thanks, I'll check it
 
  • #33

FAQ: How to determine the car's deceleration at a point along a curve?

How do you calculate the car's deceleration at a point along a curve?

The car's deceleration at a point along a curve can be calculated by using the formula a = (v^2 - u^2)/2s, where v is the final velocity, u is the initial velocity, and s is the distance traveled. This formula is based on the principle of conservation of energy.

What is the difference between deceleration and acceleration?

Deceleration is the rate at which an object's velocity decreases, while acceleration is the rate at which an object's velocity increases. Both are measured in meters per second squared (m/s^2) and can be positive or negative depending on the direction of motion.

How does the car's speed affect its deceleration on a curve?

The car's speed does not directly affect its deceleration on a curve. However, a higher speed means the car has more kinetic energy, which must be dissipated as it decelerates. This can result in a longer braking distance and a higher deceleration rate.

Can the car's deceleration be affected by external factors?

Yes, the car's deceleration can be affected by external factors such as road conditions, tire grip, and air resistance. These factors can increase or decrease the car's deceleration rate, and it is important for drivers to be aware of them to ensure safe driving.

How can the car's deceleration on a curve be measured in real-world situations?

The car's deceleration on a curve can be measured using specialized equipment such as accelerometers or by analyzing data from the car's onboard computer. In real-world situations, it can also be estimated by measuring the braking distance and the change in velocity over a certain period of time.

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