What is the required deceleration to stop a car in a given distance and time?

[sawtooth500]

Ok, this isn't a homework question, I've been out of school for quite some time but I'm having an argument with a buddy how to solve this - I know it's straight out of high school physics but it's been a loooooong time since high school physics... :)

You have a car going 44 ft/s - now you want to come to a complete stop in a distance of 132 ft and a time of 3 seconds. What is your required deceleration going to be to come to a stop?

I thought we just use d = .5 * a * t^2, working that we get a deceleration of 29.33 ft/s^2, but my buddy thinks I'm wrong - but they don't have a "right" solution either!

So am I wrong or right here, and if I am wrong, what is the correct solution to my problem?

 

[Filip Larsen]

You still posted in the homework section though

If you decelerate for 3 seconds at 29.33 ft/s² arriving at 0 ft/s at the end, what was then the speed just before you started to decelerate (hint: v = a*t)? Does this fit with a speed of 44 ft/s? If not, what does this say about how your three pieces of information fit with a constant deceleration model?

If you want more direct help perhaps you can elaborate a bit more on your setup and how you arrived at the three values 44 ft/s, 3 seconds and 132 ft.

 

[sawtooth500]

Well basically this question arose from the required deceleration to stop at a yellow light so as not to get snapped by a red light camera.

I figured car going 30 mph (44 ft/s) and a 3 second yellow. So to figure what the "decision point" would be, 44 ft/s * 3 seconds = 134 feet, so that's how I got my distance, 134 feet before the stoplight. If you are too close to the stoplight then you obviously go through the yellow, if you're too far you obviously stopped so I picked a distance where you can either choose to "go" or "not go" and then I want to back this decision up with math.

So basically you got a car going at 44 ft/s. The car then has 134 feet to stop. What is the required deceleration in ft/s^2 in order to get the car to stop in that distance from that speed?

 

[Filip Larsen]

The two "normal" kinematic equations $v = at$ and $s = \frac{1}{2}a t^2$ can be combined to give $2as = v^2$ which can be rearranged to give the acceleration

$$a = \frac{v^2}{2s}$$

and inserting this into either of the two first equation and isolating gives the time

$$t = \frac{2s}{v}$$

Using these two equations with the values 44 ft/s and 134 ft gives a deceleration around 7.2 ft/s² for a time of around 6.1 s.

Later: I just took your value of 134 ft, but I guess you really meant 3s * 44 ft/s = 132 ft as you also mentioned in your first post.

Instead of using a "single value" for the distance you can also go further and define the distance as $s = Tv$, where T is the time from yellow to red and then the two equations above reduces to

$$a = \frac{v}{2T}$$

and

$$t = 2T$$

If your car is capable of a maximum deceleration of $a_{max}$ then the maximum speed of car where it will be able to stop for red light is $v_{max} = 2Ta_{max}$.

 

[cmb]

Well basically this question arose from the required deceleration to stop at a yellow light so as not to get snapped by a red light camera.

I figured car going 30 mph (44 ft/s) and a 3 second yellow. So to figure what the "decision point" would be, 44 ft/s * 3 seconds = 134 feet, so that's how I got my distance, 134 feet before the stoplight. If you are too close to the stoplight then you obviously go through the yellow, if you're too far you obviously stopped so I picked a distance where you can either choose to "go" or "not go" and then I want to back this decision up with math.

So basically you got a car going at 44 ft/s. The car then has 134 feet to stop. What is the required deceleration in ft/s^2 in order to get the car to stop in that distance from that speed?

I'm sure you will figure out the basic equations for this, but stopping at lights is a 'different set of equations'.

This is because even if the yellow takes 3 seconds (you're lucky! Here in UK we usually get 2 seconds, even in 70 mph zones sometimes - then they want to prosecute you when you get caught out by that! Go figure!Anyhows, enough of my griping!) you have longer than 3 seconds to stop because you don't need to be stationary when the light is red, you can still be doing some decelerating.

The question then becomes - how close can I get at some particular speed that means I can still stop for the stop line.

So if your car is at 30mph, all you need is the braking distance. Here in the UK, the driver's code works this out, in feet, to be V²/20 (V in mph). So that'd be 45 feet. You can normally do much better than that, but it'd be very much an 'emergency stop' then. If you decelerate from 30mph then that gives you a stopping distance of 45 feet. If you are closer than 45 feet, then keep going. Else you can stop by declelerating at 30mph in 45 feet, which'll take longer than 3 seconds, but it doesn't matter because you'll be doing some of that deceleration after the light has gone red.

If you give yourself a second for 'thinking time' (though, supposedly, you may be anticipating a light change so you can reduce that, but let's keep it) then you have not much excuse for running a red light at 30mph if you are more than two bus lengths away from the lights when it goes amber. Within two bus lengths, keep going. More than two bus lengths, then stop. Yeah, we all hate it, me more than most because I just despise wasting all that kinetic energy. I am constantly thinking about how many Joules my car has in it, and get sad when I scrub them off pointlessly. Here in UK it costs about 10p to stop and reaccelrate a regular car from and back to 60mph. It is gallling when you stop and nothing appears from the other direction. Can't humans do better road junctions than this?

 

[cmb]

Incidentally, there is also an 'amusing' conundrum that comes out of this. In UK the laws say that you should stop for an amber if you are able to, but the offence is crossing the line on a red. This stopping-distance thing above shows that, actually, you can be in a position to stop for the amber, say 50ft, yet if you are traveling at 44ft/s then clearly you are going to be able to get across the light on the amber.

Which option do you take?

So you can cross the light on amber if you are 50' away at 30mph and keep going, whilst the law says you should stop yet has no means to enforce it.

 

[PeterO]

Ok, this isn't a homework question, I've been out of school for quite some time but I'm having an argument with a buddy how to solve this - I know it's straight out of high school physics but it's been a loooooong time since high school physics... :)

You have a car going 44 ft/s - now you want to come to a complete stop in a distance of 132 ft and a time of 3 seconds. What is your required deceleration going to be to come to a stop?

I thought we just use d = .5 * a * t^2, working that we get a deceleration of 29.33 ft/s^2, but my buddy thinks I'm wrong - but they don't have a "right" solution either!

So am I wrong or right here, and if I am wrong, what is the correct solution to my problem?

If the yellow light lasts 3 seconds, and you choose not to stop, you will travel 132' while the light is yellow.
Given that an intersection is at least 66' wide, and your car is 14' long, that means your car had better be less than 52' from the intersection if you plan to keep going, and clear the intersection before the light turns red.

If you attempt to brake, and achieve constant deceleration, your average speed while braking will be 22 ft/sec.
That means you have over 2 seconds to stop.

1 g deceleration would have you stopping in 1.3 seconds, and 1 g deceleration is pretty high for a standard road car, but you only need about 0.7g deceleration to stop before the light if doing only 30 mph.

 

[sawtooth500]

Thank you for all the help!