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Mechanics

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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Reply 1

mqb2766

Original post by Shas72

Have you sketched it? What is it you don't understand?

Reply 2

Shas72

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

Reply 3

mqb2766

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.

Reply 4

Shas72

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

Reply 5

mqb2766

Original post by Shas72

So AC= AB+BC
AB= s1/t1, BC=s2/t2
t1=t2=t

So what don't you understand?

Reply 6

DFranklin

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.

Reply 7

Shas72

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

Reply 8

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.

(edited 3 years ago)

Reply 9

Shas72

Original post by mqb2766

So A to C = s1+s2/(t1+t2)
A to B= s1/t1
B to C= s2/t2

Reply 10

mqb2766

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.

Reply 11

Shas72

Original post by mqb2766

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

Reply 12

mqb2766

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.

Reply 13

Shas72

Original post by mqb2766

Not sure what you're asking.

It says if t1=t2

Reply 14

mqb2766

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.

Reply 15

Shas72

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!

Reply 16

Shas72

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.

13(b)

Reply 17

Shas72

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 question

Reply 18

mqb2766

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.