Stopping a car before collision

  • Thread starter Thread starter hav0c
  • Start date Start date
  • Tags Tags
    Car Collision
AI Thread Summary
To stop a car before colliding with a wall, the driver can either apply brakes, turn, or apply brakes while turning. The distance needed to stop when braking is calculated as (v²)/(2μg), while turning requires a radius of (v²)/(μg). The discussion highlights that applying brakes while turning results in a weaker force component directed away from the wall, reducing the effectiveness of stopping. The analysis suggests that the optimal angle for force application significantly impacts the stopping distance. Ultimately, the most effective method for avoiding a collision depends on the specific circumstances of the situation.
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.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top