Really fast bicyclist in the park

  • Thread starter Thread starter Spartan029
  • Start date Start date
Click For Summary

Homework Help Overview

The problem involves a bicyclist's motion in a park, specifically analyzing displacement, average velocity, and average acceleration over a 24-second interval. The cyclist's initial and final positions relative to a flagpole are given, along with their respective speeds and directions.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculation of displacement using vector components and question the interpretation of angles related to direction. There is confusion about the correct representation of the cyclist's path and how to apply the results to find average velocity and acceleration.

Discussion Status

Some participants have clarified their understanding of the displacement vector and its direction. Others have provided guidance on using displacement to find average velocity and have confirmed the relationship between the average velocity vector and the angle of displacement. However, confusion remains regarding the graphical representation of average acceleration.

Contextual Notes

Participants express uncertainty about the wording of the problem and how to accurately represent the cyclist's movements in a diagram. There is an emphasis on ensuring that calculations are based on the correct reference points.

Spartan029
Messages
15
Reaction score
0
I'm stuck at part b on this problem. I've been working at it for an hour at least! lol


Homework Statement



At one instant a bicyclist is 50 m due east of a park's flagpole, going due south with a speed of 7 m/s. Then, 24 s later, the cyclist is 70 m due north of the flagpole, going due east with a speed of 20 m/s. For the cyclist in this 24 s interval, find each of the following.


(a) displacement
magnitude: ? m
direction: ? mark° north of due west

(b) average velocity
magnitude: ? m/s
direction: ? ° north of due west

(c) average acceleration
magnitude: ? m/s2
direction: ? ° north of due east

Homework Equations



\vec{}r = \sqrt{}((r\hat{i})^{2} + (r\hat{j})^{2})


The Attempt at a Solution



alright so for part (a) i used 50\hat{i} and 70\hat{j} in the equation above to get 86.023m
then inversetan (70/50) to get 54.462 degrees. i thought that 54.462 was the angle for norht of due east, but its the correct answer. It would be nice if someone could explain that to me.

then with part (b) I am completely lost, i tried plugging in the 7m/s and the 20m/s the same way as i did for the displacement but got 21.190 degrees. that's wrong and now iv been stuck forever it seems. any help would be much appreciated. thanks!
 
Physics news on Phys.org
Spartan029 said:
alright so for part (a) i used 50\hat{i} and 70\hat{j} in the equation above to get 86.023m
then inversetan (70/50) to get 54.462 degrees. i thought that 54.462 was the angle for norht of due east, but its the correct answer. It would be nice if someone could explain that to me.

Try drawing a figure with 3 points:
A point representing the flagpole.
Point #1, about 5 inches to the right (east) of the flagpole.
and
Point #2, about 7 inches above (north of) the flagpole.

Then draw an arrow from point 1 to point 2, to represent the displacement. What direction does that arrow point?

then with part (b) I am completely lost, i tried plugging in the 7m/s and the 20m/s the same way as i did for the displacement but got 21.190 degrees. that's wrong and now iv been stuck forever it seems. any help would be much appreciated. thanks!

You can use the displacement, and the elapsed time, to get average velocity.
 
Redbelly98 said:
Try drawing a figure with 3 points:
A point representing the flagpole.
Point #1, about 5 inches to the right (east) of the flagpole.
and
Point #2, about 7 inches above (north of) the flagpole.

Then draw an arrow from point 1 to point 2, to represent the displacement. What direction does that arrow point?



You can use the displacement, and the elapsed time, to get average velocity.

I see the point 1 to point 2 thing now, arrow points northwest.
the one i drew before posting here went in the opposite direction (northeast) my point 2 was 7 inches right above where my 5 inch (east) mark was, and i drew an arrow from the flagpole to that. oops. thanks for clarifying that.

so displacement divided by t= 24s would be 3.584 m/s! cool
since the Vav vector is a scalar of the displacement vector, the angle is the same, ya!?

So doing the same thing to find Aav as i did to find Vav gives me 0.883 m/s^{2}
but i don't understand how that's right. I i don't know how to represent it graphically.
 
am i allowed to bump this? :)
 
Spartan029 said:
I see the point 1 to point 2 thing now, arrow points northwest.
the one i drew before posting here went in the opposite direction (northeast) my point 2 was 7 inches right above where my 5 inch (east) mark was, and i drew an arrow from the flagpole to that. oops. thanks for clarifying that.

so displacement divided by t= 24s would be 3.584 m/s! cool
since the Vav vector is a scalar of the displacement vector, the angle is the same, ya!?

Yes.

So doing the same thing to find Aav as i did to find Vav gives me 0.883 m/s^{2}
but i don't understand how that's right. I i don't know how to represent it graphically.

Sounds right. Graphically, use the same procedure as before, except you use the velocities instead of the displacements: draw points representing 7 units south, then 20 units east.
 
Thats where I am confused. is it 7 units south and 20 east, both starting at the flagpole (origin)? or 7 south from the 50 meter mark, and 20 east from the 70m mark?
the wording on this problem is pwning me. lol
 
Spartan029 said:
Thats where I am confused. is it 7 units south and 20 east, both starting at the flagpole (origin)?

Yes.
 
cool. thanks for helping me redbelly!
 

Similar threads

  • · Replies 13 ·
Replies
13
Views
7K
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 38 ·
2
Replies
38
Views
8K
  • · Replies 8 ·
Replies
8
Views
3K
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
2
Views
4K