Uniform Circular Motion question

[devilish_wit]

Homework Statement

While learning to drive, you are in a 1 320-kg car moving at 34.0 m/s across a large, vacant, level parking lot. Suddenly you realize you are heading straight toward a brick sidewall of a large supermarket and are in danger of running into it. The pavement can exert a maximum horizontal force of 6 300 N on the car.

(1) Suppose you apply the brakes and do not turn the steering wheel. Find the minimum distance you must be from the wall to avoid a collision.

(2) If you do not brake but instead maintain constant speed and turn the steering wheel, what is the minimum distance you must be from the wall to avoid a collision?

Homework Equations

F = ma

a = V^2 / r

The Attempt at a Solution

m = 1320 kg
v = 34 m/s
F = 6300 N

(1) F = ma
6300 N = 1320kg a ---> a = 6300 N/1320kg = 4.77 m/s^2

Vf^2 = Vi^2 x 2ad
0 = (34 m/s)^2 + (2)(4.77)d
(34m/s)^2 / 2(4.77m/s^2) = d ---> d = 121.17m (This answer is correct)

(2) a = v^2 / r...

I'm not sure how to answer this second question and where to start. Any ideas?

 

[physics_pi_rate]

Hi,
We know that the mass is 1320 kg
maximum force by pavement of car is 6300N
thus maximum retardation that we can produce is = F/m= 4.77 m/s^2
using v^2=u^2 + 2as with appropriate signs we get
34^2=2⋅(4.77).s
thus s=121.17 meters for the second part let's say that while turning you're following a circular trajectory with constant speed hence it is a classical case of uniform circular motion
the centripetal force which is provided by the friction, since that is due to rubbing between the two surfaces i must say that the pavement is " kind of providing" the "turning force" which means 6300=mv^2/R where R is the radius of circular trajectory and also the minimum distance for turning if we consider length of car negligible
thus 6300=1320.(34)^2/r or R= 1320.(34)^2/6300=242.2 meters
i hope you understood my solution, i may be wrong too all suggestions are accepted. thank you

 

[devilish_wit]

Hi,
We know that the mass is 1320 kg
maximum force by pavement of car is 6300N
thus maximum retardation that we can produce is = F/m= 4.77 m/s^2
using v^2=u^2 + 2as with appropriate signs we get
34^2=2⋅(4.77).s
thus s=121.17 metersfor the second part let's say that while turning you're following a circular trajectory with constant speed hence it is a classical case of uniform circular motion
the centripetal force which is provided by the friction, since that is due to rubbing between the two surfaces i must say that the pavement is " kind of providing" the "turning force" which means 6300=mv^2/R where R is the radius of circular trajectory and also the minimum distance for turning if we consider length of car negligible
thus 6300=1320.(34)^2/r or R= 1320.(34)^2/6300=242.2 meters
i hope you understood my solution, i may be wrong too all suggestions are accepted. thank you

Yes it's correct! :)

 

[physics_pi_rate]

Yes it's correct! :)

I am glad i could be of help to you :)