Average Vector Acceleration Along a Curve and a Straight Section

  • Thread starter srekai
  • Start date
  • #1
8
0

Homework Statement


f
Capture.PNG

f

Homework Equations


##a = {v^2}{r}##
##d = vt##

The Attempt at a Solution


The way I do it is break it down into the curved and straight segment.
The bicycle is always travelling at a constant speed.
Curved portion:
distance ##\frac{\pi R^2}{4}##
acceleration ##= \frac{v^2}{R}##
time ##t = \frac{\frac{\pi R^2}{4}}{v}##

Straight portion:
distance ##R##
acceleration 0
time ##t = \frac{R}{v}##

Combining these two pieces
avg acc. = ##\frac{\frac{v^2}{R} \cdot \frac{\pi R^2}{v}}{\frac{\pi R^2}{v} +\frac{R}{v}} = \frac{\pi Rv}{\frac{\pi R^2+R}{v}} = \frac{\pi v^2}{\pi R + 1}##

As for the direction, I'm assuming it's to the right?
 

Attachments

  • Capture.PNG
    Capture.PNG
    39.9 KB · Views: 569

Answers and Replies

  • #2
FactChecker
Science Advisor
Gold Member
6,182
2,389
Since the acceleration around the arc is always pointing at the center of the circle, what would be the average direction on that part? Then there is no acceleration in the straight part, so the direction is not changed.

PS. When you use the term "right" you should be clear if you are talking about the right of the diagram or to the right of the velocity vector forward direction.
 
  • #3
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
35,915
6,644
##a = {v^2}{r}##
I assume you meant v2/r. That is the magnitude of the acceleration. You are told to average acceleration as a vector.
What is the definition of average acceleration?
distance ##\frac{\pi R^2}{4}##
That is an area, not a distance.
 
  • Like
Likes FactChecker
  • #4
PeterO
Homework Helper
2,426
48

Homework Statement


fView attachment 229300
f

Homework Equations


##a = {v^2}{r}##
##d = vt##

The Attempt at a Solution


The way I do it is break it down into the curved and straight segment.
The bicycle is always travelling at a constant speed.
Curved portion:
distance ##\frac{\pi R^2}{4}##
acceleration ##= \frac{v^2}{R}##
time ##t = \frac{\frac{\pi R^2}{4}}{v}##

Straight portion:
distance ##R##
acceleration 0
time ##t = \frac{R}{v}##

Combining these two pieces
avg acc. = ##\frac{\frac{v^2}{R} \cdot \frac{\pi R^2}{v}}{\frac{\pi R^2}{v} +\frac{R}{v}} = \frac{\pi Rv}{\frac{\pi R^2+R}{v}} = \frac{\pi v^2}{\pi R + 1}##

As for the direction, I'm assuming it's to th

Homework Statement


fView attachment 229300
f

Homework Equations


##a = {v^2}{r}##
##d = vt##

The Attempt at a Solution


The way I do it is break it down into the curved and straight segment.
The bicycle is always travelling at a constant speed.
Curved portion:
distance ##\frac{\pi R^2}{4}##
acceleration ##= \frac{v^2}{R}##
time ##t = \frac{\frac{\pi R^2}{4}}{v}##

Straight portion:
distance ##R##
acceleration 0
time ##t = \frac{R}{v}##

Combining these two pieces
avg acc. = ##\frac{\frac{v^2}{R} \cdot \frac{\pi R^2}{v}}{\frac{\pi R^2}{v} +\frac{R}{v}} = \frac{\pi Rv}{\frac{\pi R^2+R}{v}} = \frac{\pi v^2}{\pi R + 1}##

As for the direction, I'm assuming it's to the right?
Surely Average Acceleration is simply (final Velocity - initial velocity)/time taken.
In this case the subtraction is the subtraction of two vectors, and you have to work out an expression for the time interval, using speed and distance.
 
  • Like
Likes FactChecker
  • #5
20,977
4,605
Surely Average Acceleration is simply (final Velocity - initial velocity)/time taken.
In this case the subtraction is the subtraction of two vectors, and you have to work out an expression for the time interval, using speed and distance.
I think that this is what @haruspex was (more subtly) alluding to.
 

Related Threads on Average Vector Acceleration Along a Curve and a Straight Section

Replies
6
Views
1K
Replies
3
Views
6K
Replies
2
Views
2K
Replies
12
Views
2K
Replies
19
Views
987
Replies
10
Views
3K
  • Last Post
Replies
1
Views
1K
Replies
0
Views
3K
  • Last Post
Replies
19
Views
2K
Replies
8
Views
1K
Top