Math Is Fun Forum

To rearrange 3x2+3x2=6x2 to make 3x2 the subject, as you tried to do in your final step, you should have done as follows;

3x2+3x2=6x2

Take 3x2 away from both sides;

3x2=6x2-3x2 (which is 6=12-6)

Instead you took 3x2 away from the left hand side of the equation, and DIVIDED the other side by 3x2

@Phrzby Phil

“And why do we continue to call our gas pedal an accelerator once we have achieved a constant speed?”

Ah, you might be able to help me with a related physics question.
Once we’ve achieved a constant speed, and therefore we are no longer accelerating (assuming straight line travel), do we only need to keep our foot on the gas pedal to overcome friction? If there was no friction, could we take our foot off the gas and still maintain constant speed?
Also, once we’ve achieved constant speed is there no longer an unbalanced force on the vehicle?
I was initially thinking that the applied force would still be greater than the friction force and therefore the force would be unbalanced (with the gravitational force and the normal force cancelling each other out).

Thanks a lot, Bob.

I believe you! I've not quite grasped it yet, but I believe you

Just to clarify;

1. Was Google AI correct when they said 0m from the origin TYPICALLY means the object at rest? Or is even that incorrect?

2. What's the first thing that tells us, in this example, that there is no acceleration; is it the straight line slope? What else tells us?

3. What would the graph (the line) look like for an object starting at rest, accelerating from 0m/s to 5m/s, then maintaining 5m/s for several seconds; would it be curved for that first second, then straight for the remaining time?

**Respect for teaching for 37 years!   Stephen Fry (and me!) get annoyed at the lack of respect (from some) for teachers. **

Thanks, Bob.

What did you mean when you said, "...the given information does not state that u=0..."

And why did you suggest we take the initial velocity to be 5m/s? I get that if we looked at it from 1s to 3s then u = 5. But my example was from 0s to 3s.

I said that, "xf-xi = 15m-0m = 15m"

So it's position at the outset was 0m from the origin, i.e, it was at rest. Is that correct; 0m from the orgin means at rest? I suppose I could leave my house to run round the block a few times and, when I passed my home, I would be 0m from my origin but still in motion (?) In that case, how do we know what is meant by 0m from the origin?

Can we ever assume the object is at rest, when it's 0m from its origin? And if so, it has to accelerate from 0m/s to 5m/s, before it can then maintain the constant v of 5m/s (?)

NB. I'm thinking of position-time graphs with position axis going from 0m to xm, and time axis going from 0m to xs, and the line being a straight diagonal from the origin at 0m, 0s.

Thanks, Bob.

And, sorry, I didn't think I had submitted this one. I previewed it, saw all the code, and realised I don't know how to post an image on here (so I posted a thread on how to post images/photos) and thought I'd abandoned this post. Doh!

Anyway. Here is the question without all the code;

At what time does the mouse have the same position as t=0s?

I worked it out, eventually (3.5s). Did you get 4.5s?

It was the part, “...where rightwards is the positive velocity direction,” that had me puzzled.
Just to double check I’ve now got that;
Does it mean; when the velocity is positive, the displacement is to the right (and therefore, when the v is negative, the displacement is to the left)?

Thanks, Phil.

Thanks, KerimF

I had to use Google to get the formula 

v=u+at

v=+20m/s +(-10m/s^2)(5s)

v= +20m/s + -50m/s

v= +20m/s -50m/s

v= -30m/s

or;

v= 30m/s downwards

Is that correct?