Shas72
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motion from point A to point C is split into two parts. The motion from A to B has displacement s1 and takes time t1. The motion from B to C has displacement s2 and takes time t2. a Prove that if t1 = t2 the average speed from A to C is the same as the average of the speeds from A to B and from B to C .
I dont understand this question
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mqb2766
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(Original post by Shas72)
motion from point A to point C is split into two parts. The motion from A to B has displacement s1 and takes time t1. The motion from B to C has displacement s2 and takes time t2. a Prove that if t1 = t2 the average speed from A to C is the same as the average of the speeds from A to B and from B to C .
I dont understand this question
Have you sketched it? What is it you don't understand?
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Shas72
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(Original post by mqb2766)
Have you sketched it? What is it you don't understand?
Iam not able to comprehend the question. Is it talking about speed or velocity
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mqb2766
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(Original post by Shas72)
Iam not able to comprehend the question. Is it talking about speed or velocity
I'm not really sure it makes much difference. The way the question is worded, the A,B,C displacements are the distances apart. You should know the average speed formula? Could you upload a pic of the full question pls.
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Shas72
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(Original post by mqb2766)
Have you sketched it? What is it you don't understand?
So AC= AB+BC
AB= s1/t1, BC=s2/t2
t1=t2=t
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mqb2766
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(Original post by Shas72)
So AC= AB+BC
AB= s1/t1, BC=s2/t2
t1=t2=t
So what don't you understand?
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DFranklin
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(Original post by Shas72)
So AC= AB+BC
AB= s1/t1, BC=s2/t2
t1=t2=t
I don't know what you mean by AB, BC, AC here - it looks like they mean different things between the first line and the second.
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Shas72
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(Original post by DFranklin)
I don't know what you mean by AB, BC, AC here - it looks like they mean different things between the first line and the second.
I dont understand how to do this question
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mqb2766
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The first thing is to understand what they want you to prove.
* What is the average speed between A and C?
* What is the average of the two (average) speeds A->B and B->C
You have to show they're The same.

The question is about different types of means, but you don't need to know that to complete the question.
Last edited by mqb2766; 1 month ago
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Shas72
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(Original post by mqb2766)
The first thing is to understand what they want you to prove.
* What is the average speed between A and C?
* What is the average of the two (average) speeds A->B and B->C
You have to show they're The same.

The question is about different types of means, but you don't need to know that to complete the question.
So A to C = s1+s2/(t1+t2)
A to B= s1/t1
B to C= s2/t2
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mqb2766
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(Original post by Shas72)
So A to C = s1+s2/(t1+t2)
A to B= s1/t1
B to C= s2/t2
Given extra brackets and I'm guessing what the left means, then yes.
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Shas72
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(Original post by mqb2766)
The first thing is to understand what they want you to prove.
* What is the average speed between A and C?
* What is the average of the two (average) speeds A->B and B->C
You have to show they're The same.

The question is about different types of means, but you don't need to know that to complete the
t1=t2=

(Original post by mqb2766)
Given extra brackets and I'm guessing what the left means, then yes.
If i say let t1=t2=t then (s1+s2)/t =(s1+s2)/t
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mqb2766
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(Original post by Shas72)
t1=t2=


If i say let t1=t2=t then (s1+s2)/t =(s1+s2)/t
Not sure what you're asking.
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Shas72
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(Original post by mqb2766)
Not sure what you're asking.
It says if t1=t2
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mqb2766
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(Original post by Shas72)
It says if t1=t2
I know that, but you're just writing an expression.
Tbh, it's easy to complete the proof, you should be able to put the reasoning down based on what you have.
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Shas72
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(Original post by mqb2766)
Not sure what
(Original post by mqb2766)
I know that, but you're just writing an expression.
Tbh, it's easy to complete the proof, you should be able to put the reasoning down based on what you have.
Ok so i dont have to do anything more.
Thanks!
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Shas72
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(Original post by mqb2766)
I know that, but you're just writing an expression.
Tbh, it's easy to complete the proof, you should be able to put the reasoning down based on what you have.
Can you also pls help me with one more question.Name:  16022527851955637547408006915329.jpg
Views: 13
Size:  275.9 KB13(b)
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Shas72
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(Original post by mqb2766)
I know that, but you're just writing an expression.
Tbh, it's easy to complete the proof, you should be able to put the reasoning down based on what you have.
Pls help me with 14(b). I dont understand the questionName:  1602253032809864021207932768863.jpg
Views: 7
Size:  188.6 KB
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mqb2766
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Rather than just lead you through the answer, why not try and apply what you've "learnt" about average speeds to this question part. Try for 15-30 min first and post what you've tried?

Tbh this part is more interesting (not much harder) as it shows when the arithmetic mean is equal to the harmonic mean. You don't need to know this to do the question though, just use the standard average speed formula.
Last edited by mqb2766; 1 month ago
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mqb2766
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(Original post by Shas72)
Pls help me with 14(b). I dont understand the questionName:  1602253032809864021207932768863.jpg
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Size:  188.6 KB
Again p!s try and post your attempt. You learn more form trying, even if it's not successful. It's about average speed.
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