Tangential and Radial Acceleration of car around a curve

Click For Summary
SUMMARY

The discussion centers on determining the average acceleration vector of a car moving around a highway curve, specifically between two points where the car's speed doubles. The initial velocity vector is directed North, while the final velocity vector points West. Through vector analysis, participants conclude that the average acceleration is directed 26.57 degrees South of West, confirming that option (f) is the correct answer.

PREREQUISITES
  • Understanding of vector addition and subtraction
  • Knowledge of tangential and radial acceleration concepts
  • Familiarity with Cartesian coordinate systems
  • Basic principles of kinematics, particularly average acceleration
NEXT STEPS
  • Study vector decomposition in physics
  • Learn about kinematic equations and their applications
  • Explore the concepts of centripetal acceleration and its relation to circular motion
  • Investigate graphical methods for vector addition and subtraction
USEFUL FOR

Physics students, automotive engineers, and anyone interested in the dynamics of vehicles navigating curves.

123scope
Messages
4
Reaction score
0
Hi, I have a problem with this problem..
The figure (in the attachment) shows a bird's-eye view of a car going around a highway curve. As the car moves from point 1 to point 2, its speed doubles. Which vector shows the direction of the car's average acceleration between these two points?

Ok. According to my knowledge,
THe answer cannot be (g) and (b) and (a) is also unlikely
Applying the radial and tangential acceleration I guess it could be (d) and (f)
But I can't make my mind... Can anyine help me.
 

Attachments

  • 12333.jpg
    12333.jpg
    34 KB · Views: 2,004
Physics news on Phys.org
Start with the definition of average acceleration: \Delta{\vec{v}}/\Delta t.

Draw the initial velocity vector. Draw the final velocity vector. The change in velocity is the difference between those two vectors. That should enable you to choose the correct answer.

AM
 
So using a = (vf - vi) / (delta t)

I get a = (-2v i - v j) / (delta t)

So does that mean the direction of the average acceleration is in the third quadrant??
 
123scope said:
So using a = (vf - vi) / (delta t)

I get a = (-2v i - v j) / (delta t)

So does that mean the direction of the average acceleration is in the third quadrant??
Don't bother putting them on an cartesian plane - just use North/South etc. The initial velocity vector is 1 unit North and the final velocity is 2 units west. Where does the difference vector point?

AM
 
The vector diagram is as follows/
I found it to be south west. So could it be option (g)?
 

Attachments

  • 444.jpg
    444.jpg
    4 KB · Views: 991
123scope said:
The vector diagram is as follows/
I found it to be south west. So could it be option (g)?
What is the angle (south of west) that the resultant points in your diagram? What is the angle in g)? (You can measure the latter if you can't tell just by looking at it).

AM
 
sorry woops. it the answer can't be option (g). THe angle in option (g) is slightly larger.
The answer should be (f) then.
26.57 degrees south of west. Am I right?
 
123scope said:
sorry woops. it the answer can't be option (g). THe angle in option (g) is slightly larger.
The answer should be (f) then.
26.57 degrees south of west. Am I right?
Right.

AM
 

Similar threads

Replies
7
Views
4K
  • · Replies 11 ·
Replies
11
Views
2K
Replies
12
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
4
Views
4K
  • · Replies 7 ·
Replies
7
Views
6K
  • · Replies 48 ·
2
Replies
48
Views
9K
  • · Replies 9 ·
Replies
9
Views
6K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K