The coefficient of static friction between car’s tires and a level road is 0.80. If the car is to be stopped in a maximum time of 3.0 s, its maximum speed is?

Question

alekos · Answer

whats the weight of the car?

anonymous · Answer

We don't need the weight of the car ;)
First, we have to assume that the car does not use it's brakes. Then, it is only the frictional force which is doing negative work on the car, and eventually after $$t = 3.0 s$$, makes the car stop.
The car cannot have constant speed (it's speed is decreasing), so there has to be an acceleration. Newton's second law:
sum of forces = $$-R = ma$$, where R is the frictional force. Since the car is only moving in the x-direction, the forces in y-direction has to be zero. The two forces in the y-direction is the normal force N and weight G = mg. Newton's first law, with positive y-direction upwards:
$$\sum F = N-G = 0 => N = G = mg$$
Now, the frictional force R can be found, as $$R = \mu N = \mu mg$$
Apply this to Newton's second law and obtain that $$-\mu mg = ma => a = -\mu g$$
With the acceleration known, we can use the equation
$$v = v_0 - at$$, where v = 0 m/s, and $$v_0 = v_\max$$, maximum speed.
Hence, $$v_0 = at = \mu g t = 0.8 *9.81 m/s^2 * 3.0s \approx 23.5 m/s.$$

anonymous · Answer

... which is very fast :S Maybe something went wrong, check the answer.

anonymous · Answer

I also messed up on some of the plus and minus signs, but I think the answer should be right

alekos · Answer

you're absolutely right. i did the math right after i asked the question, and the mass of the car cancels out. so it's not dependent on the weight.

well done thanks Sfunn

anonymous · Answer

Ah, good to hear! Thanks :) But is it legit? Like, think of a car moving at a speed of 23.5 m/s, approx. 84,6 km/h, stopping after just three seconds due to the frictional force? The acceleration is 9.81m/s^2*0.8 = 7.848 m/s^2. The distance s covered during the slow-down is given by $$2as = v^2 - v_0^2$$ where v=0 m/s and cancels out, so that $$s = \frac {(23.5 m/s)^2} {2*7.848 m/s^2} \approx 35.2 m$$.

alekos · Answer

Yes it s legit. An average car will take around 40m to come to a halt at 80kph http://www.planetseed.com/mathsolution/braking-distances so its within the bounds of possibility that a car travelling at 84.6kph can stop after 35.2m depends on the frictional coefficient of the tyre/road surface

anonymous · Answer

Ah, now I see! I did a wrong assumption: I thought the car didn't use it's brakes but that the tires kept rolling until it eventually would come to a halt. No wonder I was confused that the acceleration was that high and that the distance driven was that short... Although, I should have known this by the fact that the static friction was given and not the kinetic friction. Thank you for helping me clear this up :)