# BR Physics 2009 Question

#### Darkskies

10+ Year Member
Hi,
I hope I'm not violating any of the rules on this forum but since I didn't find any restrictions on posting questions from review books I've decided to go ahead. In the BR physics(2009 edition) section III question 3.3b I don't understand why the correct answer is choice C instead of choice D. If the velocity factor must be squared to get the pushing distance then doesn't that mean that the distance increases more rapidly than the velocity? Choice C seems to indicate that the velocity increases at a faster pace than the distance does. Thanks for your help in advance and I am terribly sorry if posting this question is against the rules of the forum.

• To be MD

#### WorldChanger36

7+ Year Member
With that one I also went with D but it is C and here is why... From the last problem we find that KE is directly proportional to Distance because of this equation KE=W= F delta x right? So for 3.3b it just wants you to look a bit further. So setting up the two equations we find the 1/2mv^2 = F x right so we know that v and x are directly proportional. That was the easy part, then we look at v and notice it is squared there for the line has to be logrithmic but of course TBR put two log like answers on there. So how do we figure it out, well because velcity is squared we know that for every meter of displacement velcity is going to be way more. Now with graph best shows that for every meter of groud moved velocity is way higher then displacement. Only C does that. V stays way ahead of X. Going fast doesn't always mean going farther. You can cover two feet at 3x10^8 m/s or at 3 m/s and still cover two feet. That is what C is telling you. Hard thing to beat into the head but it will come with practice.

OP
D

#### Darkskies

10+ Year Member
Maybe I'm still not getting it but I thought that if the v is squared then that would mean the actual magnitude for the velocity is the square root of it since that isolates v. That would then mean that as X increases by a factor velocity increases by the square root of that factor since v^2 is not the actual velocity but the square of the velocity. I'm probably still wrong with my analysis so if you or anyone else could further elaborate it would be much appreciated.Thanks!

#### WorldChanger36

7+ Year Member
Maybe I'm still not getting it but I thought that if the v is squared then that would mean the actual magnitude for the velocity is the square root of it since that isolates v. That would then mean that as X increases by a factor velocity increases by the square root of that factor since v^2 is not the actual velocity but the square of the velocity. I'm probably still wrong with my analysis so if you or anyone else could further elaborate it would be much appreciated.Thanks!
For proportions sake lets say we get rid of everything that is not v or X in the eqation 1/2 mv^2= F X and we get V^2=X right?
If v^2 =X then v = Squarert(X) right? As v goes up X goes up as well just much slower. If you want to isolate v you have to take the square root of both sides. In other words....

SQRT(v^2)= SqRT(X) and you get v = SQRt ( X) there fore X is much smaller for every step up in v.

Write it out on paper and you will see it.

100th post sweet.... • To be MD

#### Rucap09

Bumping this, confused here.....

If V^2 = X, how would V be increasing faster than X? For example, if V goes from 2=>4, than X would go from 4=>16. In fact the question even says "If v is doubled, then x must increase by a factor of 4. This means that delta x changes more than v changes, so the graph should bend towards the x axis."

Can someone please clarify this? Thanks!

#### getsome111

V^2=X does not mean that V increases faster than X but rather the opposite. People are writing a bunch of math jargon in this post, which is technically correct, but leaves confused people with a lack of a intuitive understanding. Maybe try thinking about it this way, A constant force produces a constant acceleration. So as this force is applied the object is accelerating at the same constant rate; some increase in velocity per second. Understand that the increase in velocity is a constant, so objects traveling at higher velocities will cover more ground in the same time than objects with lower velocities if their acceleration is the same. Example; an object traveling 1 m/s with a constant acceleration 1(m/s)/s is compared to an object traveling 100m/s with the same constant acceleration 1(m/s)/s, In 1 second both objects will have increased their velocity by 1m/s, but the object that was traveling initially at 100m/s has covered much more ground than the one traveling initially at 1m/s in the same 1 second period

edit; yeah I'm aware I repeated a few things in some of the sentences, but it just felt right to do so.