Physics Help?

[ThatGermanDude]

To stop a car, first you require a certain reaction time to begin braking. Then the car slows down at a constant rate. Suppose that the total distance moved by your car during these two phases is 56.7 m when its initial speed is 83.5 km/h, and 24.4 m when the initial speed is 51.3 km/h.
(a) What is your reaction time?

(b) What is the magnitude of the deceleration?

I have been trying to set this up for hours now, but I can't get it to work. It's probably just some little thing I'm not seeing, but any help is appreciated.

[blizzard]

I would think you would use the equation

s = ut + 0.5*a*t^2

but modified in this case to:

s= u(t + tr) + 0.5*a*t^2

s is distance, u is initial velocity, t time, a acceleration, tr is reaction time.

To solve this substitute both sets of values into the equations.

Also sub the values into v= u + at and you should be able to get time in terms of acceleration which you can sub into the first set of equations and get 2 equations with only the reaction time and acceleration in them. Then just solve simultaneously.

[ThatGermanDude]

Thanks, I'll try that tomorrow. Although I tried something similar but ended up with 0=0???

[blizzard]

Sounds like you need some sleep

It should come out, looking at it now its not the easiest set of equations to solve ever.

[Brad Barker]

Originally Posted by ThatGermanDude

Thanks, I'll try that tomorrow. Although I tried something similar but ended up with 0=0???

better than 0 not being 0.

i don't particularly like the way this problem is set up at all. (and you can tell your teacher that!--"mr. so and so, some stranger on the internet who i trust is an actual physics grad student at berkeley told me that he didn't like this problem." )

i know how to solve it, though, but it ain't too pretty.

first, you want to convert all of the given velocities from km/h to m/s.

there are a buncha unknowns, and i've set up four equations. from the directions of the problem, it seems to me (though it isn't actually explicitly stated ) that the velocity that the car has during the reaction period (as well as the beginning of the braking period!) is 83.5km/h. with that in mind, we can set up three equations:

x_reaction = (83.5 km/h) (t_reaction), where "t_reaction" is, obviously enough, the length of time of the reaction period. similar definition for x_reaction.

x_brake = (83.5km/h) (t_brake) + 1/2 a (t_brake)^2, with x_brake and t_brake similarly defined.

we are given that the total distance traveled (again, the directions aren't the best) is

x_reaction + x_brake = 56.7m.

what about the acceleration? well, when the braking starts, the velocity is (presumably) 83.5km/h. when the car stops, the velocity is 0. by definition, the time during the braking process is t_brake. so you can figure out the acceleration from there.


then we have the extra initial condition and distance. i interpret this as taking place during the braking process. the kinematic equation you want to use for this one is

v^2 = (v_0)^2 + 2 a x_3. note that "a" is going to be negative, as you are braking. (you will want to write it in terms of t_brake, which you should have done in the previous two paragraphs.) the final velocity is 0; you are given the initial velocity and what i call "x_3." all in all, you can use this equation to solve for t_brake. then work your way backwards and solve for t_reaction.

[Brad Barker]

Originally Posted by blizzard

Sounds like you need some sleep

It should come out, looking at it now its not the easiest set of equations to solve ever.

the problem is not as easy as you thought, apparently!

[Phil Mailloux]

Brad, you're from California and Germandude from Houston. How come you guys are using SI units? I thought everyone in the states used the imperial system

[ThatGermanDude]

Brad, I'm not sure if you mention that in the last paragraph, but the way I understand the problem is that I'm given two scenarios: the first one where the initial velocity is 23.1944 m/s and the total distance 56.7 m and the second where the initial velocity is 14.25 m/s and the total distance 24.4 m. So I actually have two different 't_brake's?

But thank all of you for the help.

[Jared Lash]

Originally Posted by ThatGermanDude

Brad, I'm not sure if you mention that in the last paragraph, but the way I understand the problem is that I'm given two scenarios: the first one where the initial velocity is 23.1944 m/s and the total distance 56.7 m and the second where the initial velocity is 14.25 m/s and the total distance 24.4 m. So I actually have two different 't_brake's?

But thank all of you for the help.

You should have two different braking times if you are starting from two different velocities but the same deceleration rate and reaction time.

But as always, if you have two equations and two unknowns - it's substitution time.

[ThatGermanDude]

I'll play with it some more tomorrow, what I have right now is just a pain. I am substituting, but probably not the most practical things...

and I still have three unknowns? t_brake 1, t_brake 2 and either acceleration or t_reaction? Am I missing something here?

[Jared Lash]

Originally Posted by Phil Mailloux

Brad, you're from California and Germandude from Houston. How come you guys are using SI units? I thought everyone in the states used the imperial system

Everyone but scientists. But shh, don't tell anyone. We can't appear to care what anyone outside our borders thinks. It's regarded as Un-American.

Seriously, while based on the old English system, what we use are technically called US customary units.

[Brad Barker]

Originally Posted by ThatGermanDude

Brad, I'm not sure if you mention that in the last paragraph, but the way I understand the problem is that I'm given two scenarios: the first one where the initial velocity is 23.1944 m/s and the total distance 56.7 m and the second where the initial velocity is 14.25 m/s and the total distance 24.4 m. So I actually have two different 't_brake's?

But thank all of you for the help.

oh, alright. so the reaction time is about 3 seconds.

now what's weird is that the two portions of motion that you're given are woefully incomplete. there's a big chunk of motion that's missing, where the car decelerates from the 23m/s to 14m/s (i'm rounding, here). but you are given a piece of information during a portion of the braking phase, so you can work backward.

most of my work is still relevant:

x_reaction (given now!) = 23m/s t_reaction

x_brake (not given) = 23m/s t_brake + 0.5 a t_brake^2.

find "a" by noting that the final velocity is 0, the intitial velocity is necessarily 23 m/s, and the time is t_brake. (the velocities for each of the phases HAVE to match up unless the problem involves an arbitrary jump in velocity. ...as it stands, there's only an arbitrary jump in acceleration, but we can live with that--a smoothly varying acceleration would require calculus.)


then for that extra bit, it's still

v^2 = v_0^2 + 2 a x_portion, where v_0 is 14.25m/s and x_portion is 24.4m. you use "a," which itself depends on t_brake.

[blizzard]

Originally Posted by Brad Barker

the problem is not as easy as you thought, apparently!

Not really my method works, I mean obviously to solve it first everything must be in SI.

reaction time is around 0.93 seconds accelartion of -23m/s^2.


Brad looking at your method I don't quite follow what x_3 is in your last equation. I know its the displacement but wouldn't that then make it x_brake?

[ThatGermanDude]

I meant it's two completely unrelated scenarios: 1st: you're going 23.19 m/s decide to brake and brake till you stop. So you have a x_reaction, x_brake and t_brake for that scenario. Then you start going again and have to brake from a speed of 14.25 m/s to a stop. You have a different x_reaction, x_brake and t_brake. t_reaction and a are the same for both scenarios.

[Brad Barker]

Originally Posted by ThatGermanDude

I'll play with it some more tomorrow, what I have right now is just a pain. I am substituting, but probably not the most practical things...

and I still have three unknowns? t_brake 1, t_brake 2 and either acceleration or t_reaction? Am I missing something here?

ah, i see it now. the reaction time is a very quick calculation (somewhere between 2 and 3s--it's more or less handed right to you).

the acceleration, you need to solve that expression with the squares of velocities to find t_brake, which is the TOTAL AMOUNT OF TIME SPENT BRAKING. it is the amount of time to go from the speed that is the same speed you have during your reaction time to rest.

t_reaction is an unknown, but it's a straightforward, one-equation-one-unknown deal.

t_brake is an unknown. "a" is an unknown but dependent entirely on t_brake. ...and then, if you like, there is the braking distance, which you don't actually need to solve for.

[Brad Barker]

Originally Posted by blizzard

Not really my method works, I mean obviously to solve it first everything must be in SI.

reaction time is around 0.93 seconds accelartion of -23m/s^2.


Brad looking at your method I don't quite follow what x_3 is in your last equation. I know its the displacement but wouldn't that then make it x_brake?

x_3 is not x_brake. the way i defined it, x_brake is the total distance spent braking. x_3 is just some portion of that distance that was given to us.

looks like i'll have to draw some graphs so you'll know what i'm talking about...

[fdeck]

Originally Posted by Phil Mailloux

Brad, you're from California and Germandude from Houston. How come you guys are using SI units? I thought everyone in the states used the imperial system

It's actually worse than you think. In the "real world" we use both. I have seen both US and metric parts in the same product. We also have things like sheets of plywood whose thickness is given in millimeters, but the sheet size is 4 by 8 feet.

But in physics classes it's all SI. When you take an engineering license exam, you can take it in either US or metric units, and most smart people choose metric.

[blizzard]

From what I gather x_brake is the distance it takes you to come to a rest after you apply the brakes (decelerate) and is dependant on the time taken, to come to a rest and the deceleration.

The equation for final velocity squared is dependent on your initial veloctiy (given), the deceleration and distance traveled under that deceleration which is the definition of x_brake.

It could be that we both have to separate interpretations of this question.

[Brad Barker]

[blizzard]

Originally Posted by fdeck

It's actually worse than you think. In the "real world" we use both. I have seen both US and metric parts in the same product. We also have things like sheets of plywood whose thickness is given in millimeters, but the sheet size is 4 by 8 feet.

But in physics classes it's all SI. When you take an engineering license exam, you can take it in either US or metric units, and most smart people choose metric.

US measurements are the bane of my education... so illogical, on so many levels.