Math Is Fun Forum

  Yes, that's right.  It's all to do with frames of reference.  From the plane you will see me apparently moving backwards along with the ground, but I think I'm stationary.

So I'll see the ball end up at B as well.

So "The ball starts at position a, and ends up at position b. Absolutely." is the correct version.

Bob

The latitude/longitude coordinate system was devised specifically for navigating on the surface of the Earth.  You can determine your latitude by observing the angle between the Pole star and the horizontal. Latitude is harder; if you can work out the right ascension amd declination of a star and you know the time and date, you can calculate the longitude, but not much help if you're on the Moon or Mars.  No one has devised a coordinate system for the Solar System yet. 

So maybe you should make one.  World fame awaits.

Bob

hi Nick

Welcome to the forum.

I'm thinking you'd want 7 at each venue, for each course.  So person A would meet 6 others each time, that's only 24.  So that won't work.

Can I do it with more folk at some venues and less elsewhere? I suspect not but I'll go back to my spreadsheet and try.

Trivial solution that you won't want: Everyone goes to venue 1, then all on to venue 2 etc etc.

Bob

That's great!  Wow, hasn't time flown by. I seem to remember you were a lot younger when you joined.

Anyhow, it'll be a few days before I get a chance to study it properly.  I'll let you know then how I got on.  Probably have to have a few things explained (to this old timer)

Bob

Just checking: so 3 = 3 = -3   ?

Bob

Cool indeed!  Thanks for posting.

Bob

hi KerimF

Thanks for supplying a method.  I did try playing around with trig identities but didn't get there.

Sorry to be pedantic but tan(90) isn't defined ('cause you cannot play around with infinity like it's a number). But your method can be easily adapted .

Firstly

for all A except 0 and ∞.

This follows from the basic definitions of trig functions.

In a right angled triangle tanx = O/A and tan(90-x) = A/O.

Trig functions where x is not an angle in a right angled triangle are defined by a rotating unit line which ensures that any identity that is true for acute angles also holds for other angles.

So start with

This cancels down to

From which your identity follows.

Bob

That's why you need to be careful believing everything you read off the internet; especially AI.*

If you're not told the initial velocity you must not assume it is any value at all.  For some problems an object might be starting from rest but, in this problem we know it isn't.

For the whole of the 3 seconds it's travelling at a constant velocity; there's no such thing as instantaneous acceleration; so it's starting velocity must be 5 m/s.

Bob

* note: I am not a robot.  I've had to prove this loads of times by ticking boxes, even when I'm asked to tick all the squares containing a bike and one square has a tiny, tiny bit of a bike wheel trying to trick me.  I have a grade A (when A* didn't exist) in applied maths ie. mechanics and I've taught it for 37 years. Doing v = u + at problems is something I'm pretty good at.

That's exactly what I was hoping you would realise.

Well done! And that's my answer too.

Bob

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 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. 

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 Phro, Thanks for that confirmation.  Here's my method:

Bob