Braking Distance

[team1887]

Reference Problem 1-7 in the 6 Minute Transportation Solutions.

Stopping distance for a car travelling 50mph = 461ft, uncluding distance traveled during the 2.5s perception reaction time.

Another car travelling at 60mph with same distance and same friction factor, what is the required perception reaction time of this car?

Why does the solution use the equation for total stopping distance S_b = s_s + s_r instead of the SSD equation with the 1.47 and 1.075 factors? Im a little confused.

Thank you for your assistance.

 

[ptatohed]

Reference Problem 1-7 in the 6 Minute Transportation Solutions.

Stopping distance for a car travelling 50mph = 461ft, uncluding distance traveled during the 2.5s perception reaction time.

Another car travelling at 60mph with same distance and same friction factor, what is the required perception reaction time of this car?

Why does the solution use the equation for total stopping distance S_b = s_s + s_r instead of the SSD equation with the 1.47 and 1.075 factors? Im a little confused.

Thank you for your assistance.

t1,

You can refer to GDHS Eqn 3-2. SSD = (1.47)(V)(t) + (1.075)(V2/a). The "(1.47)(V)(t)" part is the sr you are referring to (distance traveled before the brakes are applied) and the (1.075)(V2/a) part is the ss you are referring to (distance traveled after brakes are applied).

First solve the SSD for car 1: SSD1 = (1.47)(V)(t) + (1.075)(V2/a) = (1.47)(50)(2.5) + (1.075)(502/11.2) = __________

Then use that distance (SSD1) for car 2 and solve for t2: SSD1 = SSD2 = (1.47)(60)(t2) + (1.075)(602/11.2); t2 = _________

Does that help?

 

[team1887]

the solution does use those factors in solving. they solve with just V*t and v^2/a. Thank you

 

[team1887]

**does not

 

[ptatohed]

Okay, so I dug deep, pulled out, and blew the dust off my old Transpo 6MS. I looked at the solution for what is problem #29 in my old edition (same as #1-7 in your edition). I had written next to the book's solution, 3 - 4 years ago when I was studying: "Too complicated! Simply use GDHS eqn 3-2 pg 113". I agree with you that the book's solution is unnecessarily frustratingly complex and cumbersomely long.

Now that I look at the problem closer, what I had written in post #2 above is not correct, sorry. I missed that the SSD1 is already given to you.

Thus, you're really solving for a1 first. So, the solution is simply:

SSD1 = (1.47)(V)(t) + (1.075)(V2/a1) => 461 = (1.47)(50)(2.5) + (1.075)(502/a1) ; a1 = 9.69 = a2

SSD1 = SSD2 = 461 = (1.47)(60)(t2) + (1.075)(602/9.69) ; t2 = 0.6986 s = 0.70 s Answer C in my edition

 

[ptatohed]

t1,

No comment? Did this not help?

 

[team1887]

Thanks for the reply. I ended up flagging and skipping over it. If I have time Ill go back to it. :/

 

[John QPE]

My take

In order for a 50 mph vehicle to require 461 feet to stop it would need a longer perception reaction time or longer decal rate, correct?? Or be on a down grade. We know perception-reaction time is 2.5 secs because that is given. It is assumed level grade, because this is not mentioned in the problem, so we assume a slower decal rate

1) solve for the decal rate of the 50 mph car

I get 9.69 ft/sec^2 when I brake the SSD equation down and solve for a

2) solve for perception-reaction time using the first decal rate

So then I solved for (t) in the second SSD equation and get 0.70 secs ..... I assumed decal rate was 9.7 secs, but now that I think more about this why would you do that??? The problem should clarify that the decal rates are the same since you are deviating from the standard 11.2 ft/s^2

Anyway ..... I got the same answer as ptatohed

My TI-36 doesn't seem to like this problem, any idea what "No sign change error" means ???

Keep coming up when I try to plug this entire equation and solve for a or t ... ihad to break it into pieces.

 

[ptatohed]

My take

In order for a 50 mph vehicle to require 461 feet to stop it would need a longer perception reaction time or longer decal rate, correct?? Or be on a down grade. We know perception-reaction time is 2.5 secs because that is given. It is assumed level grade, because this is not mentioned in the problem, so we assume a slower decal rate

1) solve for the decal rate of the 50 mph car

I get 9.69 ft/sec^2 when I brake the SSD equation down and solve for a

2) solve for perception-reaction time using the first decal rate

So then I solved for (t) in the second SSD equation and get 0.70 secs ..... I assumed decal rate was 9.7 secs, but now that I think more about this why would you do that??? The problem should clarify that the decal rates are the same since you are deviating from the standard 11.2 ft/s^2

Anyway ..... I got the same answer as ptatohed

My TI-36 doesn't seem to like this problem, any idea what "No sign change error" means ???

Keep coming up when I try to plug this entire equation and solve for a or t ... ihad to break it into pieces.

TI-36X II? TI-36X Pro? Or?

 

[John QPE]

It's the TI-36X Pro

I still haven't found an answer to this, but the Numeric Solver on this calculator did not like this problem at all.

 

[ptatohed]

..... I assumed decal rate was 9.7 secs, but now that I think more about this why would you do that??? The problem should clarify that the decal rates are the same since you are deviating from the standard 11.2 ft/s^2

I see your point. But I think they adequately state the variables that don't change (distance, friction factor), and the ones that do (reaction time, speed). Since a = f x g and we are told that f is the same for both cars (and of course we know g is the same), then a must be the same for both, and the only variable left that could change with car number 2 is the reaction time.

The stopping distance for a car traveling at 50 mph is

461 ft, including a 2.5 sec perception-reaction time. If a

car traveling at 60 mph is to stop in the same distance

with the same friction factor, what is most nearly the

required perception-reaction time?