# Uniform Circular Motion question

## 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?

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?

## Answers and Replies

Member advised not to provide complete solutions to other's homework
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 lets 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 meters

for the second part lets 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! :)

Yes it's correct! :)
I am glad i could be of help to you :)