## The motorbike and the car.

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### The motorbike and the car.

I think that this should be easy, but I am having difficulties in getting my head around it.

The problem is as follows:

A car is travelling at a constant 60km/h. In front, is a motorbike that is currently static.
The motorbike will begin to accelerate when the front of the car draws level with the rear of the motorbike.
The motorbike's maximum acceleration is such that it can attain the cars speed before the cars' rear passes the motorbikes' front.
The car is 4 metres in length, the motorbike is 2 metres in length.

The question is: Assuming constant acceleration, what is the motorbikes' 0 to 100km/h time?
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### Re: The motorbike and the car.

This isn’t really a logic puzzle, more of a math puzzle, so I’ll just provide some hints for you to think through the math:

What is an equation for the car’s position as a function of time?
What is an equation for the motorcycle’s position as a function of time?
How do these relate to each other at two particular times?
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Qaanol

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### Re: The motorbike and the car.

Despite the forum heading, the pre-header includes both Logic and Maths for this forum, hence my post here.

If I was to resolve the cars position, that would require a knowledge of time. i.e. I know that it is travelling at 60km/h, but I don't know how far it travels and therefore cannot define its duration.
Similarly, I can resolve the equestion for the motorbike to T, but I still need both A, which I am trying for and I don't know the distance.

S (or d) cannot be simply 6 metres as both motorbike and car will be moving, hence extending the distance.
I've tried to look at this a couple of ways, but feel that I keep coming to the same conclusion, that because I have 2 parameters missing (s (or d) and t), then it's not solvable.

What am I missing?
Adacore wrote:In all honesty, BigNose has been pinging me slightly with almost every post since the start of the game. But he always does - I was utterly convinced he was anti-town for most of Wizardry2 and he was the High Wizard. I just can't read him.

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### Re: The motorbike and the car.

deleted because my brain wasn't working.
Last edited by AvatarIII on Fri Dec 09, 2011 9:00 am UTC, edited 1 time in total.

AvatarIII

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### Re: The motorbike and the car.

@AvatarIII: Your numbers look inconsistent. I can walk 8.26m in 5.68s, while 2*1013m/s^2 is just insane.
And m/h as unit is a bit odd.

@BigNose: 6m is a difference between two positions. So if you look at t and a as variables, you have two equations (the relative position of both and the known final velocity of the motorbike).
Another idea is to just calculate t from a and the known final velocity and just look at a as unknown variable.
mfb

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### Re: The motorbike and the car.

Spoiler:
1.2 seconds from 0-100km/h

I brute forced it using the distance equation:

d = vt + (1/2)at2

d = distance traveled
v = initial velocity (60 km/h for the car (or 16.6667 m/s) , 0 for the motorbike)
t = time (unknown)
a = acceleration (unknown)

At t=0, d=0 is when the front bumper of the car is in line with the rear of the motorbike.

Distance traveled by the car is just v*t (16.6667t in m/s)
Distance traveled by the motorbike is .5at2

I know the upper and lower bounds for the acceleration. Lower is 0, upper is 33.3334 (which is how fast the motorbike would need to accelerate to 60km/h in 0.36 seconds, which is how long it would take the car to go 6 meters).

I just wrote a quick script for me to plug in various values for a, and have the script increment through time in .001 second increments.

For any tn, if the distance traveled by the car minus the distance traveled by the motorbike is greater than 6 meters, then the motorbike isn't accelerating fast enough, and I built in an escape in the loop.

I started with 10m/s2 (which wasn't fast enough) and just kept refining the values until I got the value of 23.1483m/s2, for the bike's acceleration.

The bike will match the car's speed of 60km/h in about .72 seconds and will have a 0-100km/h time of about 1.2 seconds.
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### Re: The motorbike and the car.

All right, I thought my hints would get you further. Here are the equations you need:

Speed of bike over time.
Spoiler:
v = at

Position of bike over time.
Spoiler:
x = ½at²

Position of bike upon reaching speed of car.
Spoiler:
x = v₁t₁ - 6m

Speed of bike upon reaching speed of car.
Spoiler:
v₁ = 60km/h = 16⅔m/s

Speed of bike upon reaching 100km/h.
Spoiler:
v₂ = 100km/h = 27⅟₉ m/s

With those it is straightforward to calculate the exact answer.
Spoiler:
t₂ = 1.1712s
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### Re: The motorbike and the car.

mfb wrote:@AvatarIII: Your numbers look inconsistent. I can walk 8.26m in 5.68s, while 2*1013m/s^2 is just insane.
And m/h as unit is a bit odd.

@BigNose: 6m is a difference between two positions. So if you look at t and a as variables, you have two equations (the relative position of both and the known final velocity of the motorbike).
Another idea is to just calculate t from a and the known final velocity and just look at a as unknown variable.

Yeah I knew it was all wrong, in fact I said that in the post I was pretty much messing around with google graphs because no one else was replying.

AvatarIII

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### Re: The motorbike and the car.

You are all using 6m as S which is incorrect.

The car is going at 60km/h which equates to 16.666m/s (v = 60000/60/60)
In order for the car to cover the minimum distance of 6m (S), it will require a time of 0.36sec. (t=S/v = 6 / 16.666)
This translates into the m'bike needing to achieve 16.666m/s (v) in 0.36secs (t) which gives a maximum aceleration of 46.29444 m/s/s (a=(v-u)/t = (16.666-0)/0.36)
As the m'bike is now moving, we have t and a to produce S, so the m'bike has now moved 2.9998m (s=ut+1/2at^2 = 1/2at^2)

Which means that the car now needs to move a further 2.9998m for its rear to meet the m'bike front.
As the m'bike is now going the same speed as the car and is accelerating, it means that the car rear will never align with the m'bike front, as the m'bike will now begin to overtake the car.

Therefore the m'bike MUST have less acceleration to converge on the cars' speed, given that the distance the car has to travel will increase and therefore the time that the m'bike has to accelerate.
As a=v/t then if v is a known constant and we are increasing t because we are increasing D, then a MUST reduce.

My thoughts at the moment suggest that some calculus maybe required on the m'bike acceleration, but I have had 3 careers since doing anything like that, hence my dilemna and now yours.
Adacore wrote:In all honesty, BigNose has been pinging me slightly with almost every post since the start of the game. But he always does - I was utterly convinced he was anti-town for most of Wizardry2 and he was the High Wizard. I just can't read him.

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### Re: The motorbike and the car.

BigNose wrote:You are all using 6m as S which is incorrect.

I don’t know what you mean by S, but 6m is the (maximum possible) distance by which the car goes further than the bike.

I gave you a correct method already. I appear to have made a numerical error in my previous calculation. Let’s try working through it algebraically.

Spoiler:
Write the distance traveled by the bike in terms of the distance traveled by the car, and in terms of the bike’s acceleration, after the same time interval. These must be equal.
tv-d=½at²

Write the speed of the bike in terms of its acceleration after that same time interval.
v=at

We now have two equations in two unknowns. Let’s solve them.

Step 1: Rewrite the second equation with t by itself.
t=v/a

Step 2: Substitute that value for t into the first equation.
v²/a-d=½v²/a

Step 3: Solve that last equation for a.
a=½v²/d

Step 4: Write equation for time s to reach arbitrary speed w.
as=w

Step 5: Solve for time s.
s = w/a

Step 6: Plug in equation for a.
s = w/(½v²/d)

Step 7: Simplify.
s = 2dw/v²

Step 8: Plug in known values.
s = 2(6m)(100km/h)/(60km/h)²

Step 9: Simplify.
s = 20m/(60 km/h) = 1mh/3km = 3600ms/3000m = 1.2s
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Qaanol

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### Re: The motorbike and the car.

can someone tell me what is wrong with my graphs?

why am I getting 2.4s and not 1.2s? I know 1.2s is right, but I'm asking why acceleration needs to be doubled to get the right answer?
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AvatarIII

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### Re: The motorbike and the car.

The acceleration is ~23.15m/s^2. No need to double it.
If you want to compare the distance, note that s=1/2 a t^2 and not a t^2.
mfb

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### Re: The motorbike and the car.

mfb wrote:The acceleration is ~23.15m/s^2. No need to double it.
If you want to compare the distance, note that s=1/2 a t^2 and not a t^2.

OK that's what I mean, why does acceleration need to be halved

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### Re: The motorbike and the car.

Lets work this back then.
If the m'bike achieves 100km/h in just 1.2 secs, then its accerlation is:
v=u+at which is a=v/t which is a= 100*1000/60/60=27.7778m/s /1.2 secs which has a value of 23.148m/s/s

As this is a constant, then we can work out the time taken to get to 60km/h and ofcourse, the distance travelled by the m'bike.
v=u+at which is t=v/a which is 16.6667/23.148 = 0.72 secs
s=ut+1/2at^2 which is 1/2at^2 which is 0.5*23.148*.72*.72 = 6m

Which is to say that the m'bike must travel 6m to reach the same speed as the car, which also happens to be the same distance the car has to travel as a minimum.
Coincidence?
Which is why I suspect your calculations.
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### Re: The motorbike and the car.

I think we are all agree on the 1.2s and 23.148ms-2 numbers. the figure for when the bike's front and car's back are level is definitely 6m, even though it doesn't look like it should be going by my graph. the calculations say 6m

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### Re: The motorbike and the car.

BigNose wrote:Coincidence?

Nope. Without doing any calculations at all aside from one simple addition (car length plus bike length), I could have told you the bike would travel 6 meters in its acceleration to the car's speed just by considering the problem from a different reference frame. From the reference frame where the car is at rest, the bike is speeding backwards and decelerating to a stop, and will go from its initial speed of 60 km/h to a stop in the combined length of the two vehicles - 6 meters. The magnitude of the bike's acceleration is the same in both frames, and 0-to-X and X-to-0 cases are symmetric so they take the same time and distance, so by that symmetry the bike's distance traveled in the original reference frame will also be 6 meters.

That 6m distance is an extra indication that the calculation is correct, and is even an alternative route to the answer. The car must travel precisely 6m more to bring the specified front/rear alignment about, so 12m = 60 km/h * t. One easy division and some unit conversions, and you get t = .72 seconds. Plug that into d = 1/2 at2, getting 6 = .5 * a * .722, and divide to get a = 23.148 m/s2. For another double check, the speed of the bike at that moment is 23.148 * .72 = 16.6667 m/s or 60 km/h, matching the car's speed exactly as it should.

And yes, this gives 1.2 seconds as the time for the bike to reach 100 km/h.
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### Re: The motorbike and the car.

In which case, thank you and apologies for the doubt.
Adacore wrote:In all honesty, BigNose has been pinging me slightly with almost every post since the start of the game. But he always does - I was utterly convinced he was anti-town for most of Wizardry2 and he was the High Wizard. I just can't read him.

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### Re: The motorbike and the car.

Nice observation douglasm.

AvatarIII wrote:OK that's what I mean, why does acceleration need to be halved

Area of a triangle.

Spoiler:
Given constant acceleration a, the graph of acceleration versus time is a horizontal line.

Assuming 0 initial velocity, final velocity is the time-integral of acceleration. With constant acceleration, the area under the acceleration curve is just a rectangle of height a and base t, so velocity at time t is v=at. The graph of velocity versus time is thus a straight line of slope a.

Assuming 0 initial position, final position is the time-integral of velocity. With constant acceleration and 0 initial velocity, the area under the velocity curve is just a triangle of height at and base t, so position at time t is ½at².
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### Re: The motorbike and the car.

Since this has already been solved, I will attempt to give as simple a solution as possible. No calculator required.

Start with the 100 kmph problem. We can do this by simply stretching the distance of the original problem. The car is now traveling 100 kmph and at the same acceleration, the distance increases with the square of final velocity and the car gains 10*100/60 m, or 50/3 m instead of the original 6 m. Because the bike is accelerating linearly, we know that it has half the average speed of the car. So the total distance traveled by the car must be twice the distance gained by the car, or 100/3 m. At 100000 m/h how long does it take to travel 100/3 m? 100/300000 hours. To turn that into seconds, (60*60)/3000 seconds, or 1.2 seconds.

But it can all boil down to t = (4m+2m)*2 /60000 m/hr * 100km/hr /60 km/hr * 60 s/min * 60 min/hr
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### Re: The motorbike and the car.

I think the simplest way of looking at it is as follows. In the time taken for the motorbike to accelerate to 60km/h, the car goes 6m further than the motorbike. The average speed of the bike over that period is 30km/h and so the car is 30km/h faster on average, so will travel 6m further than the bike in the time it would take to travel 6m at 30km/h. The bike would therefore reach 100km/h in the time it would take to travel 10m at 30km/h, which is 36000/30000=1.2s.
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### Re: The motorbike and the car.

Guys, we don't have one crucial ingredient: its maximum speed, without that, we cannot calculate its 0 to 100Km/h time.
Last edited by tomtom2357 on Wed Jan 25, 2012 1:30 am UTC, edited 1 time in total.
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### Re: The motorbike and the car.

tomtom2357 wrote:Guys, we don't have I crucial ingredient: its maximum speed, without that, we cannot calculate its 0 to 100Km/h time.

In accordance to comic 169, a solution like that would get someone's hand chopped of. Not sure whose though.

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### Re: The motorbike and the car.

tomtom2357 wrote:Guys, we don't have I crucial ingredient: its maximum speed, without that, we cannot calculate its 0 to 100Km/h time.

Tomtom, we are told to assume constant acceleration. With that assumption, there is no maximum speed. In reality, of course, acceleration is not constant, and there is a maximum speed, but for the purposes of the puzzle, you're supposed to ignore reality.
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### Re: The motorbike and the car.

This is just a problem of getting the right equations.

I'm going to work with velocities measured in meters per minute so the numbers are easy (1000 m/min). Also, the reference point of the car will be at its back and the reference point for the bike will be at its front. So at t=0, the car is 6 meters behind the bike. Their positions with respect to time:

s(car) = -6 + 1000t
s(bike) = 1/2 at^2

Also note the bike's velocity with respect to time:

v(bike) = at

We are told that when s(car) = s(bike), it will be the case that v(car) = v(bike) = 1000 m/min
We have two equations and to unknowns.

-6 + 1000t = 1/2 at^2 and at = 1000
So -6 + 1000t = 1/2 (1000)t (by substituting at = 1000)

This is a linear equation, and we solve t = 6/500 min (or 0.72 sec)

We now solve for a: a(6/500) = 1000
so a = 500 000/6 m/min^2 = 500/6 km/min^2

As for it's 0 to 100 time, we want
at = 100/60 km/min
(500/6) t = 10/6
t = 1/50 minutes, or 1.2 sec to go from 0 to 100.
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### Re: The motorbike and the car.

Technically, it just says before the tires match up the second time. Any ridiculously large acceleration value will satisfy the original question, what we really need is the minimum (which obviously, has been found.)
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