00:01
So in this problem, we are going to do some kinematics with stopping distance, and we're given a bunch of information.
00:07
We're told that the speed limit of the road, as it is now, is 55 miles per hour, and that there is a visibility of 155 feet.
00:20
So this is basically going to be in one situation our stopping distance, because you can't react to stop to something unless you see it.
00:30
We then give in some bounds for the coefficient of friction, that the coefficient of rolling friction, so we'll call it muar, is somewhere between 0 .842 and 0 .941, and that the coefficient of skidding friction, when the tires are locked up and grinding across the road, somewhere between 0 .550 and 0 .550 and 0 .741.
01:13
We also know that the cars could be the mass of the car, could be anywhere from 1130 pounds to 8 ,940 pounds for a bigger truck.
01:29
So to determine stopping distance, we need to do some kinematics.
01:33
And that starts with a free value diagram.
01:36
So we have some car, which we can just put as a block, or even more generally, we can put it as just a point.
01:43
And what are the forces acting on it? well, firstly, we have the force of gravity, which is equal to mass times gravity, so it's weight.
01:53
And then so that the car doesn't fall through the road, there is some normal force pointed upwards based on the contact from the car on the road.
02:06
Then the car is going to stop.
02:09
So let's say we have our x and y axis.
02:18
If velocity, if the initial velocity was greater than zero, in other words, if it was in the positive direction, then if it was stopping, if the velocity was therefore like that, to stop, we need an acceleration in the opposite direction, which would be negative, and therefore the force would be in the negative direction.
02:36
So we can put that here, which is our friction force, because that is how the stopping happens based on the wheels, which is our friction force, which is equal to the coefficient of friction.
02:49
Friction times the normal force.
02:53
So now what we want to do is develop a equation that has excursion in it, and to do that, we're going to use newton's second law, where by newton's second law, the sum of forces is equal to mass times exclamation.
03:07
We're going to start in the x direction here, and say that the sum of forces, the only force here is the friction force.
03:16
So we're going to say that the coefficient friction times n is equal to mass times acceleration.
03:22
And in general, we know the friction is going to be back but for this case, what we're going to do is write it in the positive direction, and the sign will come out because we know acceleration should be negative.
03:34
We can leave it like that.
03:35
We'll leave it like that.
03:36
We just know that acceleration is negative because this should be negative based on the sign convention, but we know that this itself is also negative.
03:47
So we're going to drop that sign for this part.
03:51
So we know mass.
03:53
We know we're going to be able to find exclamation through kinematics.
03:56
But what would n be? we need to go to the y direction then.
04:00
We can consider each separately, so the sum of forces in the y is equal to mass times acceleration.
04:06
But because the car is staying on the road, the acceleration in the y is equal to zero, and so they all equal to zero.
04:12
And that results in negative mg plus n equal to zero, or n equals mg.
04:21
We can then plug that in, and that's not g, that is a.
04:30
And then we see we have m on both sides of the equation.
04:32
They cancel out.
04:34
And so in the x direction, we have simply the coefficient friction times the acceleration of gravity here on earth is equal to the acceleration in the x direction, which is what we want.
04:46
Now importantly, mass is not in this equation.
04:49
So there's one piece of data we're given about the mass of the cars.
04:53
We don't care about.
04:54
This is going to be true for any mass of car assuming the fricking coefficients are true.
05:01
So what we're going to do then is fine with the stopping distances.
05:05
That we can use a kinematic equation where the let's not do this in red let's do this in black where the final velocity squared is equal to the initial velocity squared plus two times the acceleration times the position minus the initial position so we're going to set up our car at some initial point x equals to zero and it stops over some distance to a final x it has v -not going to speed v.
05:38
This is stopping, though, so we know that the final speed is equal to zero.
05:43
We know that x not is equal to zero like we already set, because that's where we set our axes.
05:47
And then this x here is the stopping distance.
05:52
So we can make the simplifications to our equation.
05:55
We know the final velocity is equal to zero.
05:58
Initial position is equal to zero because we can set the axis wherever we want.
06:02
So we can then solve this for our stopping distance x, and that is over 2a, is equal to x.
06:16
Now we expect a positive distance, but we have this negative sign here, but that's okay.
06:21
We remember that acceleration is going to be negative, so that works out.
06:26
So now we can then use this equation to find the minimum and maximum distances.
06:32
First thing, though, is we need to figure out how tovert miles per hour to feet per second, because we want to go into the base units.
06:38
So for a 55, or for any speed, which is miles per hour, we know that there are 5 ,200 ,000, 180 feet in a mile, and then in one hour, there are 3 ,600 seconds, which is this 60 minutes times 60 seconds.
07:09
And doing this dimensional analysis, we see that the units cancel that we don't want, and we'll have its feet per second in the base units like we want.
07:16
So if we have any miles per hour and multiply it by 5 ,280 over 3 ,600, then we're getting a few per second, which is what we want.
07:28
Now we can look at how we can get the minimum maximum distances.
07:31
Well, our distance is what we're solving for.
07:35
We know the velocity.
07:36
So the only thing that can change is the acceleration.
07:40
And while we're here, we can also plug in when acceleration is, because we know that it is equal to mu -g.
07:52
And we also put the negative sign in here.
07:56
So the only thing we can really change is mu because g is going to be constant.
08:00
So the minimum distance, if we go back to equation, we're dividing by a number.
08:05
So if mu is smaller, we're dividing...