Stuck in a question from exam -- Car decelerating to avoid hitting an obstacle

[PeroK]

View attachment 301663
Okay, after you told me all what i need to do I'm still not even sure if i did it correctly, is it correct ?? please correct me if I'm wrong.

That initial distance of $5m$ seems to have been included three times! Twice would be too many, but three times is excessive, I would say!

I think your calculation for $X_2$ is correct. And, given that $X_2 > 90m$ you needn't have worried about the reaction time. Although, you added $5m$ to it.

 

[Delta2]

You did it almost correct, only mistake is that you "double count" the distance d=5m that the car travels during the 0.15second.

 

[PeroK]

You did it almost correct, only mistake is that you "double count" the distance d=5m that the car travels during the 0.15second.

Treble count, even!

 

[Delta2]

Er no its double lol, He counts it once in step 2 (where he shouldnt) and once in step 3 (where he should).

 

[shuumi]

That initial distance of $5m$ seems to have been included three times! Twice would be too many, but three times is excessive, I would say!

I think your calculation for $X_2$ is correct. And, given that $X_2 > 90m$ you needn't have worried about the reaction time. Although, you added $5m$ to it.

the second part is correct, I'm confused about the 5m thing, what is wrong about it? since the acceleration is not included, the only way to do it is by using the basic method that we know: V = d/t, do i have to use an another method? Or i have choose the wrong numbers ?

 

[shuumi]

You did it almost correct, only mistake is that you "double count" the distance d=5m that the car travels during the 0.15second.

what do you mean by double count ? Is like using it two times which was unnecessary ??

 

[PeroK]

the second part is correct, I'm confused about the 5m thing, what is wrong about it? since the acceleration is not included, the only way to do it is by using the basic method that we know: V = d/t, do i have to use an another method? Or i have choose the wrong numbers ?

The $5m$ is correct, but you added it three times instead of once.

 

[shuumi]

The $5m$ is correct, but you added it three times instead of once.

In the second step, do i have to remove it and replace it to zero ? so in this case, does that mean that the Xo after the acceleration is 0, because i though it's was 5m (if like we talked in general). Is this the wrong thing about 5m ?

 

[PeroK]

In the second step, do i have to remove it and replace it to zero ? so in this case, does that mean that the Xo after the acceleration is 0, because i though it's was 5m (if like we talked in general). Is this the wrong thing about 5m ?

You calculate $5m$ distance traveled before the brakes are applied. Then you calculate the distance traveled to decelerate to a stop. Then you add them.

You can't add the $5m$ twice. It's knowing that that allows you to organise your equations. Not the other way round.

 

[PeroK]

Er no its double lol, He counts it once in step 2 (where he shouldnt) and once in step 3 (where he should).

You're right. I thought we had got up to $109 \ m$ but I misread the $2$ as a $9$.

 

[shuumi]

You calculate $5m$ distance traveled before the brakes are applied. Then you calculate the distance traveled to decelerate to a stop. Then you add them.

You can't add the $5m$ twice. It's knowing that that allows you to organise your equations. Not the other way round.

Oh so they're like separated equations, i think i understand this exercise now.

 

[shuumi]

Thank you so much for everyone that helped me and gives me the advices about the right strategy to do and the solution, i really appreciate that from every one of you, and i still have some other exercises that i 'm confused about, i will try my best to do them, if i needed help, i will come back here again, so please take care of me !

 

[erobz]

In the second step, do i have to remove it and replace it to zero ? so in this case, does that mean that the Xo after the acceleration is 0, because i though it's was 5m (if like we talked in general). Is this the wrong thing about 5m ?

The equations of motion change at the end of the first time interval.

for $x_0 \to x_1$ the velocity is constant $v_o = v_1$:

$$ \Delta x_{0 \to 1} = v_ot $$

for $x_1 \to x_2$ the velocity is changing from $v_1$ to $v_2 = 0$:

$$ v_2^2 - v_1^2 = 2 a \Delta x_{1 \to 2} $$

The total distance traveled is then:

$$ \Delta X = \Delta x_{0 \to 1} + \Delta x_{1 \to 2} = v_o t + \frac{-v_1^2}{2a} $$

 

[PeroK]

Oh so they're like separated equations, i think i understand this exercise now.

Here's an idea. Do the problem with zero reaction time and see what you get.

The answer with $5m$ reaction distance should only be $5m$ longer. If it's $10m$ longer, then you've double counted the $5m$.

 

[shuumi]

Here's an idea. Do the problem with zero reaction time and see what you get.

The answer with $5m$ reaction distance should only be $5m$ longer. If it's $10m$ longer, then you've double counted the $5m$.

If we do 0 as a time of reaction so this is not a problem about a car crushing isn't ?

 

[Delta2]

No it will still remain a problem about a car crushing, but let's just say that the driver is a former F1 champion and has zero reaction time. What should be the $X_1$ with zero reaction time?

 

[Delta2]

Reaction time is the time that elapses from the moment the driver sees the obstacle, till the moment that he presses the break pedal.

 

[shuumi]

No it will still remain a problem about a car crushing, but let's just say that the driver is a former F1 champion and has zero reaction time. What should be the $X_1$ with zero reaction time?

I didn't really understand wdym by a former F1 champion, but if he has a 0 reaction time, then X1 will be igual to 0 since there is no time of reaction.

 

[Delta2]

Correct X1=0 in this case.

 

[shuumi]

Correct X1=0 in this case.

Then we will calculate the other unknown values in a normal way without the third stage of the addition of the two distances since Xo is 0. i think i understand those types of exercises now.

 

[bob012345]

Then we will calculate the other unknown values in a normal way without the third stage of the addition of the two distances since Xo is 0. i think i understand those types of exercises now.

I think it would be easier to first compute the distance traveled before the driver could react which you did at 5m then simply say the obstacle is now 85 m away. Simply ask can the car going 33.33m/s stop in 85 m at an deceleration of -6m/s^2?

Using $v_f^2 - v_0^2 = 2 a \Delta {x}$ we have with $v_f = 0$ $\Delta {x} = \large \frac{-33.33^2}{2(-6) m/s^2} = 92.6m$ which is greater than the 85m left to go so it crashes.

You could also ask how fast is the car going when it crashes?

 

[Delta2]

Yes the exact time and speed at the moment of collision would be another nice question for this problem.

 

[PeroK]

Here's how I would have done this problem. First, I would have calculated the vehicle's stopping distance, based on the deceleration of $-6 \ m/s^2$ and the initial speed of $33.3 = \frac{100}{3} \ m/s$. As the final speed is $v = 0$ and using $v = u + at$ we have:
$$0 = \frac{100}{3} \ m/s + (-6 \ m/s^2)t$$ which gives
$$t = \frac{100}{18} = \frac{50}{9} \ s$$I would then use that the average speed for deceleration to rest is half the initial speed. So, $v_a = \frac{50}{3} \ m/s$. And the distance traveled is $$s = v_at = \frac{50}{3}\frac{50}{9} \ m = \frac{2500}{27} \ m \approx 92.6 \ m$$So, we can see that the vehicle needs more than $90m$ to stop, even without the reaction time.

Note that an alternative approach is to use $v^2 - u^2 = 2as$. And, with $v = 0$, $a = -6 \ m/s^2$ and $u = \frac{100}{3} \ m/s$ we have:
$$s = \frac{100^2}{9 \times 12} \ m = \frac{2500}{27} \ m \approx 92.6 \ m$$Which is the same as before.

Although we don't have to calculate the additional distance covered during the reaction time, we know that it is $5m$. That means the total stopping distance (reaction plus braking) is $97.6 \ m$.

Note that a better problem would have the obstacle $95 \ m$ ahead, as that would require us to calculate the reaction distance.

 

[bob012345]

i think i understand those types of exercises now.

That is our goal. Good luck on your exam!

 

[Nik_2213]

A tad rusty on the math, but I dredged my 'v=u+ft' and 's=ut+0.5ft^2', set to work.
First, convert that innocuous kph to m/s. I lived by mph & fps, but same implacable math applies.
Yup, at that speed, you lose a car's length before the 'Reed Alert' kicks in.
Assuming you brake then swear, you're still a limo 'long'.
You and your passengers are wearing seat-belts, yes ??

( Air-bags are good, but deflate just in time for next vehicle's impact to hurl you through screen... )

Which is why the careful driver must anticipate rather than simply react, while limiting speed to worst-case scenario...

( Eyes on stalks. Scan traffic flow for ijits: Remember, they're all out to get you... )

Funny story for you:
One of our truck drivers had a lapse on motorway, plowed 40-ton 'artic' into standing traffic. Wasn't 'speeding', on the phone or anything, just sort of blinked. So, all of us insured to drive on company business, with truck or car, had to do an urgent road-safety course. Finale was a poster-sized hazard-spotting 'toon. Think 'Hectic, Very Cluttered Side-street'. Ring all the horrors. Pass was 16/20.
There was much hilarity when I handed in my sheet with 24 'potentials' ringed.
Until our 'Health & Safety' boss took a closer look, decided mine were all valid, made a few calls.
Seems 'Corporate' mistakenly circulated only the first page of 'spots'. The second page had my extra four.
So, instead of pass being 16/20, it was 20/24, and a lot of people had to do the course again...

 

[haruspex]

Air-bags are good, but deflate just in time for next vehicle's impact to hurl you through screen

Presumably the next impact is coming from behind, so the risk is whiplash, not a faceplant.

 

[Nik_2213]

For a 'minor' impact, yes, but a 'major' impact that rams your vehicle a lot further into the wrack is a different story...

 

[haruspex]

For a 'minor' impact, yes, but a 'major' impact that rams your vehicle a lot further into the wrack is a different story...

Yes, interesting point.