A Question for Math Freaks

[levymetal]

Hey everyone,

I have a question that I need some help with. I used to be pretty good at maths, but I have no idea how to go about answering this. I was wondering if you have 2 different cars, that do 0-100 in x amount of time, would they cover the exact same amount of distance? I've been thinking but I just can't get my head around it.

[stocky]

i would think that the slower car would cover a longer distance

[levymetal]

haha, sorry i meant 2 cars that do 0-100 in the same amount of time (x)

[graemevb]

Short answer no, the distances travelled can be different.

Long answer, consider two cases, one extreme
Car A: that accelerates at a constant rate of 10km/h per sec, it will take 10 seconds to reach 100km/h. The average speed is 50km/h(13.9m/s), distance travelled =139m

Car B: accelerates at 2km/h for 9 secs, average speed is 9km/h(2.5m/s) for 9secs, for a distance of 22.5m.
The car then needs to go from 18km/h to 100km/h in 1 sec, which is an acceleration of 82km/h for 1 sec or an average speed of (100+18)/2 = 59km/h(16.4m/s) for that 1 sec, in which it would cover 16.4m. Total dist = 22.5 + 16.4m = 38.9m.

[pow3rslave]

if they both do 0-100 in the same amount of time, then assuming they accelerate at the same rate they will be neck and neck indefinitely.
i think even if one accelerated differently, like gear ratios made one slower at low speeds, but faster than the other higher up and they also didn't get to 100, one might be ahead, but assuming they go to 100 they'll be at the same place.

[nick3110]

well they are going at the same acceleration, they must cover the same distance.. But there are alot of variables that come along with this question, tyres, weather, condition of car etc.. but on the simple side they would travel the same distance..

[graemevb]

well they are going at the same acceleration, they must cover the same distance....

That would be true, if the rate of acceleration were to be constant, but it is not one of the constraints of the question.

[levymetal]

ah, so basically that means that time, distance travelled, and speed are all directly connected right?

but can't you do the same time on 1/4 or 1/8 mile at different speeds?

[levymetal]

alright i just read graemevbs comment. let me reflect on it for a sec.

[graemevb]

ah, so basically that means that time, distance travelled, and speed are all directly connected right?

but can't you do the same time on 1/4 or 1/8 mile at different speeds?

Yes you can have different speeds across the line but the same time, speed across the finish line only indicates how fast you were going at that point.

Think of a 100m foot race where one person overtakes a slower runner right on the finish line, they take the same time and have both run the same distance, but one is going faster than the other.

[Crossy_014]

graemevb is correct

think about it

[vlv8vic]

That would be true, if the rate of acceleration were to be constant, but it is not one of the constraints of the question.

i think the question is ambiguous.

If they cover the same distance in the same time then you'd need to assign a time or distance to measure at, eg 100m. At which point both cars are there, making them equal. While that might be true, the story at an earlier point might be different. One car might have sat at the line for 5 seconds spinning the wheels while the other accelerated gently to the 100m mark, where the speed of the other car eventually made up the ground lost due to traction problems.

There are too many pieces of info missing to get a definitive answer.

[graemevb]

i think the question is ambiguous.

If they cover the same distance in the same time then you'd need to assign a time or distance to measure at, eg 100m. At which point both cars are there, making them equal. While that might be true, the story at an earlier point might be different. One car might have sat at the line for 5 seconds spinning the wheels while the other accelerated gently to the 100m mark, where the speed of the other car eventually made up the ground lost due to traction problems.

There are too many pieces of info missing to get a definitive answer.

There is nothing wrong with the question, it simply asks if the final speed and time taken are the same, will the distance covered have to be the same? How is that ambiguous? No it doesnt depend on untangibles.
A definitive answer is no, the distances can be different, depending on which how they are accelerated to that speed within the given time, Ive given you A and B scenarios, did you read them?
While the case B scenario is extreme, it illustrates the point, real cars dont accelerate at constant rates,(even if they have have great traction and cvt) the rate of acceleration drops away as speed increases, how much depends on the aerodynamics of the vehicle.

[vlv8vic]

^^^ i read it as 0-100 metres. I get where you are coming from.

[Not_An_Abba_Fan]

It would come down to average acceleration up to the given speed. If one car accelerates faster initially, and the other had a faster acceleration a bit later, if both had the same average acceleration, then the distance covered would have to be the same.

Take the 1/4 mile example, a faster top end wold result in a faster terminal speed than a car that accelerated faster on the lower end then slowed down. Both vehicles may achieve the same ET but they would have different terminal speeds. Thus, the car with the faster TS would reach a higher given speed first, covering a shorter distance. But then if it was a slower speed that needed to be reached, say 50km/h, then the car with the slower TS would hit that speed first, having a faster initial acceleration.

[graemevb]

It would come down to average acceleration up to the given speed. If one car accelerates faster initially, and the other had a faster acceleration a bit later, if both had the same average acceleration, then the distance covered would have to be the same.
.

No that is incorrect.

As defined in the problem, both cars attain the same final speed in the same given time, this means the average acceleration has to be the same. They will only cover the same distance if their average speed is the same.

If you look at the car A and car B scenario I gave in post #4, the average acceleration is indeed the same(has to be the same): both cars go from 0 to 100km/h in 10 secs so the average accel = 10km/h per sec, but the distances covered are NOT the same. For car B the average accel can also be worked out as: 2km/h for 9 sec and 82 km/h for 1 sec = 2 x .9 + 82 x .1 = 10km/h per sec

Perhaps another way to understand the phenomenon is to look at it graphically and plot the speed versus time. The distance travelled is equal to the area between the curve plotted and the horizontal axis, perhaps see: Determining the Area on a velocity-time Graph
The car that accelerates at a high rate and then reduces to a lower rate will cover a much greater distance than a car that does vice versa, even if the average acceleration is the same
As can be seen on the graph, the area between the curve and horizontal axis is less for car B than car A, car A goes further than car B

[Philthy]

Alright...I'm going for an extreme example on this, but will keep it as simple as I can.

Assume car A accelerates from 0-80 in 1 second, then takes 9 seconds to get from 80-100. He is going to have 9 seconds travelling at an average speed of 90kph or 25m/s so must travel 25*9 = 225m in the 80-100 stage. Add on 1 second of an average speed of 40kph (11.1m/s) for the 0-80 stage, and you get 225+11 = 236m for the total 10 sec 0-100 run

Car B accelerates from 0-80 in 9 seconds...average speed of 40k's for 9 sec = 100m travelled. 80-100 in 1 sec, average of 90kph = 25m. Car B only travels 125m in the 10 seconds it takes him to get to 100k's.

Damnit I just read up and saw Graemevb already had a good example...I don't care, I still wrote this out so I'll post it anyway

[1991_Vn2nV]

Short answer no, the distances travelled can be different.

Long answer, consider two cases, one extreme
Car A: that accelerates at a constant rate of 10km/h per sec, it will take 10 seconds to reach 100km/h. The average speed is 50km/h(13.9m/s), distance travelled =139m

Car B: accelerates at 2km/h for 9 secs, average speed is 9km/h(2.5m/s) for 9secs, for a distance of 22.5m.
The car then needs to go from 18km/h to 100km/h in 1 sec, which is an acceleration of 82km/h for 1 sec or an average speed of (100+18)/2 = 59km/h(16.4m/s) for that 1 sec, in which it would cover 16.4m. Total dist = 22.5 + 16.4m = 38.9m.

I think everyone should read this before posting, as its a perfect example.

I swear if you explained to some people that 2+2=4 they would still try and argue.

[vlv8vic]

I think everyone should read this before posting, as its a perfect example.

I swear if you explained to some people that 2+2=4 they would still try and argue.

i don't believe anyone's arguing - just a few getting it wrong!!!

[Philthy]

No that is incorrect.

As defined in the problem, both cars attain the same final speed in the same given time, this means the average acceleration has to be the same. They will only cover the same distance if their average speed is the same.

If you look at the car A and car B scenario I gave in post #4, the average acceleration is indeed the same(has to be the same): both cars go from 0 to 100km/h in 10 secs so the average accel = 10km/h per sec, but the distances covered are NOT the same. For car B the average accel can also be worked out as: 2km/h for 9 sec and 82 km/h for 1 sec = 2 x .9 + 82 x .1 = 10km/h per sec

Perhaps another way to understand the phenomenon is to look at it graphically and plot the speed versus time. The distance travelled is equal to the area between the curve plotted and the horizontal axis, perhaps see: Determining the Area on a velocity-time Graph
The car that accelerates at a high rate and then reduces to a lower rate will cover a much greater distance than a car that does vice versa, even if the average acceleration is the same
As can be seen on the graph, the area between the curve and horizontal axis is less for car B than car A, car A goes further than car B

Surely this graph settles it once and for all. Distance is the area under the speed vs time graph, so if you don't see it straight away count the squares under each of the lines.
If anyone still disagrees it can get a lot more technical than looking at a graph

[Tsunamix]

Short answer no, the distances travelled can be different.

Long answer, consider two cases, one extreme
Car A: that accelerates at a constant rate of 10km/h per sec, it will take 10 seconds to reach 100km/h. The average speed is 50km/h(13.9m/s), distance travelled =139m

Car B: accelerates at 2km/h for 9 secs, average speed is 9km/h(2.5m/s) for 9secs, for a distance of 22.5m.
The car then needs to go from 18km/h to 100km/h in 1 sec, which is an acceleration of 82km/h for 1 sec or an average speed of (100+18)/2 = 59km/h(16.4m/s) for that 1 sec, in which it would cover 16.4m. Total dist = 22.5 + 16.4m = 38.9m.

Umm... Average speeds assume that you are travelling at that speed for that period of time, which is clearly not the case as ytour speed is changing all the time.

Think about it this way. Gravity is a force of 9/8 metres per second, per second. 9.8 m/s/s which is a measure of it's acceleration.

Drop a weight of 1kg out a window and in one second it will have travelled 9.8m and be travelling at 9.8 m/s. At 2 seconds it will have travelled 29.4 m (9.8 distance from the end of second one, plus another second travelling at 9.8 + 9.8 m/s ). At 3 seconds its 58.8m and so on.

The whole thing comes down to the old F=MA equation. Force = Mass x Acceleration. Your force and mass can still vary, yet result in identical acceleration. 10hp=1Kgx10 units of acceleration. 10Hp = 10kgx1 unit of acceleration.

In this case you ask if 2 differing masses can accelerate at 2 different rates and still end up physically at the same point at the same time. The answer is yes, it just requires different forces. Those forces can vary every 20'th of a second if they like, it is still possible. 100kg accelerated at by a force of xx nm for 3 seconds and yy nm for 3 seconds, will be ion exactly the same place at the same time as a 100kg mass accelerated by a force of (xx+yy)/2 for 6 seconds.

[Philthy]

Umm... Average speeds assume that you are travelling at that speed for that period of time, which is clearly not the case as ytour speed is changing all the time.

Think about it this way. Gravity is a force of 9/8 metres per second, per second. 9.8 m/s/s which is a measure of it's acceleration.

Drop a weight of 1kg out a window and in one second it will have travelled 9.8m and be travelling at 9.8 m/s. At 2 seconds it will have travelled 29.4 m (9.8 distance from the end of second one, plus another second travelling at 9.8 + 9.8 m/s ). At 3 seconds its 58.8m and so on.

The whole thing comes down to the old F=MA equation. Force = Mass x Acceleration. Your force and mass can still vary, yet result in identical acceleration. 10hp=1Kgx10 units of acceleration. 10Hp = 10kgx1 unit of acceleration.

In this case you ask if 2 differing masses can accelerate at 2 different rates and still end up physically at the same point at the same time. The answer is yes, it just requires different forces. Those forces can vary every 20'th of a second if they like, it is still possible. 100kg accelerated at by a force of xx nm for 3 seconds and yy nm for 3 seconds, will be ion exactly the same place at the same time as a 100kg mass accelerated by a force of (xx+yy)/2 for 6 seconds.

It might just be me, but I'm pretty sure all that had nothing to do with 2 cars with identical 0-100 times. Would you prefer the term "mean" speed to be used?

[levymetal]

yep that graph just blew my mind, makes total sense now. thanks guys

[Tsunamix]

It might just be me, but I'm pretty sure all that had nothing to do with 2 cars with identical 0-100 times. Would you prefer the term "mean" speed to be used?

Yeah mean speed would have made sense, I thought you were ignoring the whole accelerative process, and suggesting calculations based on instant transitions.

Bah it's friday arvo and I have been looking at pedantic contract specifications for too long

[graemevb]

Umm... Average speeds assume that you are travelling at that speed for that period of time, which is clearly not the case as ytour speed is changing all the time.
.

I dont know what the point of your post is, are you saying Im wrong??
Umm, average speed = distance travelled divided by time taken, there is no simpler way of putting it. Perhaps you are confusing constant speed and average speed.

If in a journey the car is constantly accelerating then it only does the average speed for a brief instant of time half way through the journey, otherwise its going less than the average speed for half the journey and more than it for the other half, think about it.

I read the first line of your next bit of the physics lesson, but oh so wrong!

Drop a weight of 1kg out a window and in one second it will have travelled 9.8m and be travelling at 9.8 m/s.
.

If you drop any object than after 1 sec it will be travelling at 9.8m/s, but it will have only travelled half that distance 4.9m, not 9.8m in 1 sec. The average speed for the 1st second being 4.9m/s....does that help.

Ill stop there, Im sure the rest is similarly flawed.