Can any kind smart anon help me with this? And explain their

>>724117496
Twelve

>>724118213
How? I'm trying to learn so I don't have to resort to 4chan again. Thanks

niggers

>>724117496

where's the bus coming from?

the front or back?

>>724119342
How

>>724117496
Seems like you've translated this and haven't really conveyed all information to us properly. Is the bus meeting and the car wants to pass the truck before coliding with the bus? Where is it coming from? Try drawing it up, always helps.

>>724119852

thought the same thing.

he is a nigger with words.

>>724119755
Not sure. It says:

>the car and truck are traveling down a highway and he bus is approaching him on the other side of the road.

>>724117496
take your homework back to >>>/hm/

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>>724119939
>>the car and truck are traveling down a highway and he bus is approaching him on the other side of the road.

who the fuck writes this shit?

>>724119852

Here's the problem again, verbatim.

A car and a truck are both traveling 30 m/s down a highway. The car is 10m behind the truck and the truck is 20m long. The car is 5m long. The driver of the car sees that a bus is approaching him on the other side of the road traveling at a rate of 25 m/s.The driver decides to pass the truck by accelerating constantly at 3m/s^2 until he passes the truck and pulls into his travel lane 20m in front of the truck. Determine the minimum distance that the car needs between the front of the car and the front of the approaching bus when the car begins to pass the truck.

that is a long fucking car

>>724120065
Take your moms ashes and put them back in the vase

>>724120189

car is only 5 meters.

its traveling at 108,000 miles per hour though. so its heavy as fuck.

>>724120304
5 meters is long as fuck. thats over 16 feat
u mean a limousine?

>>724120065
Not homework, problem I'm trying to work out my textbook so I can better my understanding.

Not me, not this edgy
>>724120261

>>724120138
So when the car goes back into his travelling lane, is the distance of 20m between the back or front of the car?

>>724120304
That's 108 000km/h or 67 068mph roughly mach 87 I think the 5 metre length is to account for the rocket engine propelling it

>>724120512

kek. nobody understands the language of the problem.

>>724120512
>pulls into his travel lane 20 meters in front of the truck.

Is this what you're asking? There are only 3 vehicles involved in this question. A car, truck, and bus.

If someone can just tell me what to do to find "minimum distance" that would be helpful too.

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>>724120654
>5 metre length is to account for the rocket engine propelling it

implying theres room for fuel for such rocket to propel a car to 108K mph.

>>724120741

you need about tree fiddi m/s.

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>>724117496
Anyone?

>>724120859
Put some respek on my damn problem

Ok why not...
1. you calculate the time the car needs for the maneuver. You assume starting velocity 0, acceleration as described, distance as described
2. you calculate the distance the bus can travel during this time relatively to the truck (add both velocities)
3. you add the distance between the starting point of the car to the end point of the maneuver
4. enjoy winz

>>724120741
Ok, so you know that car and truck are starting with the same constant speed v=30m^/s.

Add a=3m/s^2 for the car.

Calculate time needed for the car to travel 55m more than if it had held it's constant speed. This is assuming the distance they asked for is 20m between front of the truck and back of the car. If front to front do same for 50m.

Last you need to calculate how far something travelling the same distance at 25m/s would be.

That's it if I've interpreted this correct.

>>724117496
basically the question is:

how far does the bus have to be for the car to overtake the truck without smashing into said bus (approaching from the opposite direction)

so first we determine how far the car needs to travel relatively to the truck:

so the distance is 5(car's length)+10(distance between truck and car)+20(truck length) =35m

we call this s1
now we know that the truck and car start out at the same speed and the car after which the car accelerates at 3 meters per sec squared

so from accelleration we need to derive the formulas for speed, acc and distance
s = v * t
s= .5 at²

a = v / t

using these 2 formulas will allow you to get all your variables and solve the questions

1) det. distance for car to travel (relative)
2) det. time to travel this distance based on accelleration
3) det. how far the bus will travel in this same time (use relative accelleration/speed here because bus and car are approaching each other)
4.) check anwser to see if you calculated relative speed or actual speed and check what the question required (if actual add the original travel speed of car to get real distance rather then relative

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>>724120741
Also, if you were to solve this at an exam you should draw something like this, but adjusted to your problem.

>>724121387
Your s1 is incorrect, you've forgotten the 20m distance the car travels past the truck before changing lanes.

>>724117496
1. Equations of Motion
v = u + at
s = ut + 1/2at^2
s = 1/2(u + v)t
v^2 = u^2 + 2as
s = vt - 1/2at^2

2. Do your own homework

>>724120797
>implying theres room for fuel for such rocket to propel a car to 108K mph

Plutonium for 1.21 gigawatts will do it.

>>724121141
>>724121151
Thanks

>>724121387
You are a gentleman and a scholar thanks.

>>724121645
Thanks. I'm trying, can only do what you understand though right

>>724121549
Op here, so everything else is correct, the s1 is just equal to 55m

The answer is 12, honestly not that difficult.

>>724121387
What about the car's drafting/aspiration then slowing down when begining to pass the truck?

>>724121975
Can you show the process please

>>724121883
>>724121549
yeah my bad it's 55


so you get smthing like

s=½at² => t=~ 6
with s=55m., a=3m/s²

and s=vt => s=~333 m.
with v=55m/s, t= 6.05


>>724121975
you sure that a bunch of vehicles traveling just under 100km/h can pass each other in 12 meters? even tho the actual truck is allrdy longer then that, seems unlikely u should check anwsers if they make sense.

>>724121975
How? I got something way off

>>724121975
55 for back of car to front of truck. 50 if front to front. Try drawing it up for clarity.

>>724122526
Where did the 6 come from?

>>724122642
sqrt((55*2)/3)

He is breaking t free from s=.5at^2

>>724122526
>>724122642
>>724122642
>>724122642

maths

s=½at² -> 55=½ 3 t² -> t²= 55/1.5 -> t= sqr(36.6\) -> t= 6.05 ish seconds

>>724122526
Ah ok I'm starting to follow. Do you mind working it all out. It will help me out a lot better

>>724123029
25*sqrt((55*2)/3)=answer

>>724123029
Oh well might be askew, considering the constant speeds of the car and truck in the beginning you might want to add 30*55 as well. 1.8km then, maybe reasonable, they are travelling fairly slow all of them. Does your book give you the correct answer?

>>724123127
Why because the bus is at a constant velocity of 30m/s so I want to multiply that by the time it takes the car to clear s1?
>>724123443

Lol no not in the addenudim. A bit weird since this is a professor made textbook. If it was, I wouldn't be asking the board.

>>724123029
lol i literally did work it all out xd

so if you get the first part here >>724122985

then you do the same with the other formula (which is s=vt ) but this time you use your time calculated in step one, to solve the other unknown in this formula (being s0


so you know the bus is headed towards the car and they are traveling towards each other.
one at a 30m/s speed and the other at 25 m/s

this means that relatively (watching from the car for example) the approach speed will seem like 30m/s + 25m/s because the bus is headed towards you, and you are allready moving at a speed towards the bus.
this gives us 55 m/s for the speed v in this formula
since we just determined that the car needs about 6 sec to overtake the truck we know that that's the time that we need to use to calculate how far that bus will travel in the mean time
therefore we use the t we just calculated.

this gives us a formula with one unknown, that we can solve, the unknown being s (distance traveled by the bus)
why>> well u need 6secs to overtake so the bus needs to be atleast that, or more seconds away from passing you (otherwise car->pancake)

now just solve the formula

s=vt -> s=55*6.05 -> s=~333m

>>724123596
I keep realizing I've been missing stuff in the question. We are not looking for the distance between front of bus and car when the car is in its starting position behind the truck, but when it first starts to overtake it. But the distance the bus travels should be added on as well since they are all meeting.

>>724123737
I got 151.25m as minimum distance

>>724124088
Anyone get the same?

>>724123893

not really the point of these types of questions is to make you realise distance speed and acceleration are primitive/derivate functions of one another towards time, but even more so it teaches you to pick your reference frame.

eg. it;s a lot faster and easier to calculate this using the cars reference frame.
in other words calculating this from the cars vieuwpoint is way easier the from a passersby viewpoint, because you don't need to, all you need to know is how far the bus is from the car, and not anything about where it actually is

>>724124088
that's a factor of exactly ½ so either you or me missed a ½ factor in one of our derivate formulas
make sure you dont use avg. speed here but ad the two velocity's of bus and car, they are traveling in direct opposite directions

>> i might have made some error smwhere, but cba to doublecheck this again, u can scroll up ot check mah maths :P

>>724123737
But when looking at the the car relative to the bus you haven't taken the cars acceleration in consideration? I don't think you can just add their velocities if both of them aren't constant.

>>724124611
Aye, I'm not OP, just trying to remember my old classes. Think I've been messing the question up too much trying to interpret it.

Distance the truck travels during the time it takes for the car to completely overtake it as required plus the distance the bus travels in the same time should be correct. This gives me approximately 330m, so you're probably right.

>>724124704
>>724124704
so i guess that's it then

pulling his apart would yield :

t=6 still that was correct

but now we get the bus which is s=vt =35*6
car s= .½at²+ vt (we add the distance due to initial speed and the distance due to acceleration)
you will see the equation is now 2nd order including a factor for speed and acceleration
car s= ½*3*6² +20 * 6

so thts like 175 is + 210 is giving us a new anwser of around 385m.

which makes sense since this number is higher, then our precious anwser, cause we took acceleration into account

>>724125959
Of course! We have s=.5*3*6^2+30*6 for the car and 25*6 for the bus though, still gives about 385m.