+0  
 
0
351
13
avatar+1828 

 

 

 

physics
xvxvxv  Oct 9, 2014

Best Answer 

 #11
avatar+26412 
+10

For anyone uncomfortable with the determination of the limiting values as δt goes to zero in the above, here are graphical representations of those limits:

 

limits as dt goes to zero

 

.

Alan  Oct 14, 2014
Sort: 

13+0 Answers

 #1
avatar+26412 
+5

Q3.3

Acceleration is a vector quantity, so it is non-zero if either the magnitude or direction (or both) of the velocity is changing.  Point C is the only marked point where neither the magnitude nor direction of velocity is changing, so the answer is B.

 

Q3.7

I'm not sure what the symbols above v and a mean, but, ignoring these:

There is a component of acceleration in the i-direction which is making the object deviate from its path in the j-direction, so the object is on a curved path.  The component of acceleration in the j-direction is in the opposite direction to the direction of the velocity, so tending to slow the object, so the answer is C.

 

.

Alan  Oct 9, 2014
 #2
avatar+1828 
0

Thank you, but what about this 

 

xvxvxv  Oct 9, 2014
 #3
avatar+1828 
0

Alan, this is my interpret for Q3.3 why at point c the direction doesn't change . 

 

At C the object moving along a straight line so its acceleration is tangent to the path and parallel to the velocity 

 

 I don't know if it is correct or no 

Also I want your own interpret to learn from you 

xvxvxv  Oct 14, 2014
 #4
avatar+91505 
+5

Alan,   may I ask a question too?

In Q3.7   What if the i acceleration was huge, would it still have to be slowing down since the j accel is negative?

Isn't the i component of any consequnce at all to slowing down or speeding up if the the original velocity in the i direction is 0?

I'm interested in Q3.8 as well.   (there is an extra ^2 in the i velocity component that shouldn't be there)

It is curved

It is getting faster in the i direction and slower in the j direction.

How do you work more than that out?

Melody  Oct 14, 2014
 #5
avatar+26412 
+10

Q3.3   For an object to be accelerating it must either be changing speed or direction or both.  The question states the object is moving at constant speed, so it can only be accelerating when it is changing direction.  At point C it is not changing direction, it is travelling in a straight line.  So, at C it is changing neither speed nor direction; therefore it is a point of zero acceleration.  At the other points its direction is changing, hence it is accelerating, hence they are not points of zero acceleration.

 

Q3.7 and 3.8 I'll do a more complete analysis with a diagram later. 

Alan  Oct 14, 2014
 #6
avatar+91505 
0

Thanks Alan

I understood 3.3, it was the other 2 I was questioning.    So I look forward to your extended explanation.  

Melody  Oct 14, 2014
 #7
avatar+26412 
+5

For question 3.7:

 

Not quite right - see below.

 

.

Alan  Oct 14, 2014
 #8
avatar+26412 
+5

In error - see below.

 

 

.

Alan  Oct 14, 2014
 #9
avatar+91505 
+5

Thanks Alan,

You've given me lots to think about.  Thank you.   

Melody  Oct 14, 2014
 #10
avatar+26412 
+10

I made some errors in my previous analysis.  A more general approach is derived and used here:

 

General derivation

Q3.7 and Q3.8 results

So, in question 3.7 the result is a reduction in speed and in question 3.8 an increase in speed.

 

(I've used δa for acceleration, but should probably have just used a.  I had δv = aδt in mind .  This doesn't affect the analysis though.)

.

Alan  Oct 14, 2014
 #11
avatar+26412 
+10
Best Answer

For anyone uncomfortable with the determination of the limiting values as δt goes to zero in the above, here are graphical representations of those limits:

 

limits as dt goes to zero

 

.

Alan  Oct 14, 2014
 #12
avatar+91505 
0

Hi Alan,

I understand at least some of what you have done.

I am wondering about your derivation.

I can follow it alright but how did you do the simplification.     (Where it actually says simplify)

Can you do that by hand or do you need to use a calculator such as Mathcad or Wolfram|Alpha?

Thank you.:)

Melody  Oct 18, 2014
 #13
avatar+26412 
+5

Here are the steps that lead to the simplification (note that I have replaced the δa's by a's in the following)

 

derivation pt1

derivation pt2

 

 

.

Alan  Oct 18, 2014

5 Online Users

avatar
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details