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Whoops, I did mean 3.5. I just counted squares along the graph and miscounted. Hey! I'm a mathematician; can't be bothered with all that counting business.
The velocity rightwards makes no sense at all since the velocity axis is up/down.
Here's my interpretation of the graph.
A mouse wants to cross the road and is waiting at the kerb. A big lorry thunders along and the mouse is so frightened it accelerates rapidly away from the road; then decelerates at the same magnitude; coming to rest after two seconds. As the lorry has now passed the mouse continues the same acceleration back to the road; at t=3 it maintains the constant velocity it has reached of magnitude 5 m/s and 0.5 s later arrives back at the kerb. It continues with that velocity and crosses the road without mishap; arriving at the other side at t=6.
To achieve this interpretation the positive velocity is upwards and represents velocities away] from the road. The road is 12.5 metres wide.
Odd question but maybe in mouse kinetics it makes more sense. Mice do spend a lot of their time running away from danger so they would take 'away' as the positive direction.
Bob
hi Nancycastro25
Welcome to the forum.
This thread is a very old one and the last post was in 2012. Good to contribute (that's what this forum is for) but the last poster hasn't posted again since then ![]()
Bob
hi paulb203
Here's the graph:

Below your descriptive paragraph is some code that isn't displaying usefully for me. Also there are bits here that don't make sense for me.
rightwards is the positive velocity direction.
But v is the up axis.
And what is this mouse doing? It starts with zero velocity and over a period of one second accelerates to 5 m/s. Then, in another second it decelerates back to zero at (2,0). It continues to accelerate negatively (ie accelerate but in the opposite direction) for another second and then continues at a steady speed for 3 more seconds.
By using 'area under the graph = distance travelled' I can see it has travelled 5 metres after 2 seconds. But then the areas are negative so it's coming back from its original direction. -5 metres occurs at 4.5 seconds when it is back where it started?? Then it continues in that negative direction until 6 seconds by which time it has travelled 5 x 2.5 metres in the negative direction. Has it crossed the road? Only if it was pointing the wrong way to start with and ran the wrong way until 4.5 seconds.
Bob
hi Alison,
Something strange has happened with the formating of your post. On my laptop there are too many new line characters so it's hard to work out what you're saying. Does it look ok to you?
If so, how are you viewing the forum?
Or are you an AI bot?
Bob
hi Phro, Thanks for that confirmation. Here's my method:
Bob
hi ryanjaberan
Welcome to the forum.
This problem has taken me back 53 or so years. Had to rework some formulas.
I think the answer is
If that's correct then I'll post my method. If it's not, break it to me gently. ![]()
Bob
Yes it is.
Wow! You've heard of Spike Milligan. My Dad had that record on a bakelight 78 rpm disc. Surely you're too young?
Bob
Yes it does. A question ought to start by defining which direction is positive. I think we had this in a previous post where an object was being thrown upwards. You can take up or down as the positive direction but this needs to be consistent throughout the question. eg. if 'up' is positive then gravity is -g.
Bob
That's a fair description. I'll stick to across and up in the future.
Bob
The correct terms for the across and up axes are abscissa and ordinate. Descartes thought up the coordinate system and used x and y, so those have kind of stuck. The page where I found the definitions of abscissa and ordinate even used x and y to explain which is which ![]()
So it is good practice to always label each axis and it's perfectly ok to use your own symbols where the context suggests it (such as t across for time).
It is also usual to make x the independent variable (ie. the one whose values we can choose) and y the dependent variable (ie. the one whose values depend on a formula involving x).
But usual, not compulsory.
In 3D coordinates x is usually shown going right, y going back into the page, and z going up.
Is it confusing? Yes, probably, but using x for time would be worse I think. A graph is just a way to make a picture out of a mathematical model and usually helps. Choosing the right variables will help if the grapher says what and why.
Saying at the start what symbols you're going to use is essential. In logic there are several ways to indicate logic operators such as AND, OR, IMPLIES etc. And it is common to use a dot to replace a multiplication symbol to avoid confusion with a variable x. In an Argand diagram the up axis labels have an 'i' to indicate an 'imaginary' number, but some folk use 'j'.
Bob
Not all authorities agree that a scalar must be positive.
Eg. Temperature is a scalar and on some scales can be negative.
Bob
You say 'initial' and 'final' so I'm assuming you're using v = u + at.
I think v and u here are vectors. It's what I was taught .... velocity a vector, speed a scalar. But I tried google. The first 17 hits all had v as a vector. None as a scalar.
Strictly there should be some indication such as an overline or bold print,
but nobody seems to bother. But I guess that's why Khan thinks scalar.
In practice it'll come out in the calculation, so don't worry too much about the distinction.
Bob
hi Mageurna
Welcome back to the forum. ![]()
Bob
distance and time are defined as fundamental units https://en.wikiversity.org/wiki/Fundamental_units
Speed is a derived unit.
The ancients certainly had the concept. Mariners, for example, would throw a piece of wood overboard to estimate the speed of their ship. And I'm sure many horses were sold taking account of their speed.
Speed is a rate of change (of distance with respect to time) but it doesn't seem like they had the idea of miles per hour (using their own unit for distance).
So maybe Galileo was the first to do this.
Bob
6 feet is scalar but deep implies a direction ie downwards so that makes it a vector.
If you were standing in a trench, 6 feet deep, and you fired an arrow upwards you'd have to use the equations of motion to determine what happens next. Those equations use vector quantities, ( except time).
Bob
I can see there's a lion approaching. Would you just like to run in a random direction or would it be helpful to know which direction is best? ![]()
My ship is wanting to intercept another ship on the ocean. It is also moving. How do I plot a course so that we meet at some place ahead? Shall I just choose a new speed or might it also be useful to consider the direction of travel?
Three ropes are tied together in a knot. Two others are ready to pull on their ropes and want me to pull also so that, together, our forces exactly balance. Do you think it matters which directions we each choose for our pull?
Bob
The forum moved to a new server in March '21 (I'll call this moment the 'change'). Since then avatar uploads have not been possible. But members with avatars from before the change still have them so the code controlling them in a members profile still works. Just the upload is faulty.
I've searched all the admin options available to me and found nothing to turn this back on. It's odd that Ganesh and I can get to the upload page. Is this because we are admin, or just that we already have avatars?
I'm reluctant to try it for myself in case I lose the one I have. But I can log in on an unused account so I might try an experiment. Also might try making a new admin just so I can test if uploads will work.
Looks like there's some code missing from the new forum but I don't have acccess to that.
LATER EDIT: Promoting a user to admin didn't work. The link led to the upload screen but the file wouldn't upload.
EVEN LATER EDIT: I've tried to change the avatar for a very old member who last posted in 2005 and would be banned today due to the nature of their posts. That member had an avatar. An attempt at using the link gave the usual error message.
YET ANOTHER EDIT: I've taken the risk and tried to change my own avatar. Couldn't upload one. Conclusion: No one can upload at present whatever their status.
Bob
hi Ganesh,
The link works for me as 'bob' but not as 'alter ego' , a member. It'll be interesting to find out if it works for ktesla39
Bob
![]()
Actually it's worse than that. I should have calculated the area of CEG. So that calculation is completely wrong.
It should be:
0.5 times EC^2 times root(3)/2 which for my diagram comes out as 0.88
Still working on an analytic solution.
Bob
hi
Has your account been hacked? I've deleted a post that I'm sure wasn't made by you and, as a precaution, changed the password on the account.
Please reply here as a new member so we can sort this out (ie. get you back in control of the account)
Bob
I'm assuming you used Geogebra and moved a point around until you reached a minimum. I did that using Sketchpad and got this:

I feel that there ought to be an analytical way to do this, so I'm working on that.
Bob
Is their answer good enough? I'd say no .... a lot has been left out. Your proof is much better. ![]()
Bob
Yes, for my random choice for a and b. You can try it with your own choices.
Bob
I've found a way.
step 1. consider a parabola with F = (0,a) and D is y = -a
If (x,y) is the general point on this parabola
Set a = 1/4 and we have y = x^2, so this well known curve is a parabola.
step 2. let a = 1/(4A) => y = Ax^2 is also a parabola.
step 3. Transform this by vector (0,d) => y = Ax^2 + d is a parabola.
step 4. consider y = ax^2 + bx + c
let the bracketed term be d/a
By the vector transform X = x + b/(2a) this becomes y = AX^2 + d and so is also a parabola.
Thus all quadratics of the form y = ax^2 + bx + c are parabolas.
Bob
I'm working on a proof that all curves of the form y = ax^2 + bx + c are parabolas.
Progress so far:
y = x^2 fairly easy.
y = ax^2 not too much harder.
y = ax^2 + bx
y = ax^2 + bx + c follows easily by translation.
Bob