Stopping a car before collision

  • Thread starter Thread starter hav0c
  • Start date Start date
  • Tags Tags
    Car Collision
Click For Summary
SUMMARY

The discussion focuses on the physics of stopping a car before a collision with a wall, considering factors such as velocity (v), mass (M), and the coefficients of static and kinetic friction (μ). The key equations include the distance traveled when brakes are applied, calculated as (v²)/(2μg), and the radius of a turn, given by (v²)/(μg). The analysis reveals that applying brakes while turning results in a reduced force magnitude, leading to less effective deceleration compared to braking alone. The discussion emphasizes the importance of understanding force components and their impact on stopping distance and turning radius.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with friction coefficients (static and kinetic)
  • Basic knowledge of circular motion and angular velocity
  • Proficiency in solving equations of motion
NEXT STEPS
  • Study the principles of friction in vehicle dynamics
  • Learn about the effects of braking distance on different surfaces
  • Explore the concept of centripetal force in turning vehicles
  • Investigate advanced braking techniques and their physics
USEFUL FOR

Automotive engineers, physics students, driving instructors, and anyone interested in vehicle safety and dynamics will benefit from this discussion.

hav0c
Messages
58
Reaction score
0

Homework Statement



A person driving a car suddenly sees a wall .At that instant velocity of the car is v. Coefficient of static and kinetic friction is μ. The mass of the car is M. To avoid the driver should ...(acc. due to gravity=g)
a)apply breaks
b)turn (the car shouldn't slip while turning)
c)apply breaks while turning (the car won't topple)

Homework Equations


v=rω
f=ma
v=d(s)/d(t)
... other equations of motion
(ignore actual scenario of pumping the brakes)

The Attempt at a Solution


In case the driver applies the brakes the distance traveled is simply (v2)/(2μg).
In case the driver turns the radius of the turn will be (v2)/(μg))
How will i work out the case of applying breaks and turning ?
 
Physics news on Phys.org
Keeping maximum Force magnitude applied at some (fixed) angle from -v ...
you know the result for 0° is half the result for 90°.
Suppose you try it at 45° (then a∙r=-v∙v makes r become a function of remaining KE ... too bad).
The Force component pointing away from the wall is 71%, weaker than the simple braking case (100%),
so the Work done by that component while traveling to the wall must be less, at any other angle.
This means that the forward component of velocity is changed less, if F applied at any non-180 angle.
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
4
Views
3K
  • · Replies 9 ·
Replies
9
Views
5K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 1 ·
Replies
1
Views
8K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K