#16**+5 **

*1- you made the net force constant because the track is fricitionless ? *

*2- how did you determine that the acceleration is positive ?*

Identify the forces on the glider. Without friction (or air drag etc) the only forces are gravity and the normal reaction force of the track on the glider.

The reaction force balances the component of the gravitational force normal to the track (I'm assuming the glider stays on the track!).

The component of the gravitational force parallel to the track is the *only* force acting to change the speed of the glider, and, since gravity and the slope of the track are both constant, the force is constant. This force acts in the downhill direction which is in the positive direction. The acceleration is therefore in this direction.

**Acceleration doesn't necessarily lead to an increase in speed**. Because the glider starts with an initial speed **up** the track the positive acceleration first acts to slow it down, before bringing it to a halt and **then** increasing its speed back downhill.

.

Alan
Oct 8, 2014

#1**+5 **

The first one is C, velocity is negative because the object is moving in a negative direction.

The acceleration is also negative because the object is covering more ground in each equal time period and it is speeding up in a negative direction.

Melody
Sep 28, 2014

#2**+5 **

The last one is A. The object is slowing down at a decreasing rate.

the velocity is pos at the begining and near zero at the end so it must be slowing down.

The gradient of the curve is slowly decreasing so the rate is decreasing (the curve is concave up)

The middle question I am not so sure about. The question does not talk about the acceleration at all.

I don't think enough info is given.

Maybe Alan will come back and help us out here.

Melody
Sep 28, 2014

#3**0 **

Thanks alot but can you tell me the best way to know the behaviour of the slope specially with concave functions ! not the straight line functions

xvxvxv
Sep 28, 2014

#4**0 **

a tangent is a line that just touches the curve.

When I say that the slope and is decreasing with time I mean the gradient of the tangent of f(t) is decreasing. It is becoming less steep as the time increases.

Lets look at these velocity verses time graphs.

For A, C and E the velocities are increaing with time.

A is incresing uniformly so acceleration is a positive constant.

C is increasing at an increasing rate (the tangent to the curve is becoming steeper, the curve is concave up)

Hence, for C the acceleration is positive and increasing.

E is increasing at a decreasing rate ( the tangent to the curve is becoming less steep, the curve is concave down)

Hence, for E the acceleration is positive and decreasing, at the end of the curve the acceleration has become 0.

For D,and F the velocities are decreasing with time. (Although they are both traveling in a positive direction)

D is decreasing at a decreasing rate (the tangent to the curve is becoming less steep, the curve is concave up)

F is decreasing at an increasing rate (the tangent to the curve is becoming more steep. The curve is concave down)

B is speeding up but in a negative direction. The acceleration is constant and negative

G, H and I have zero acceleration. G is stop so velocity is zero, H has positive velocity and I has negative velocity.

I think these are all okay but **Alan, could you please check what I have said.**

Melody
Sep 28, 2014

#5**0 **

What an explanaition ... I can't believe my eyes Its very wonderful You put all the states in this comment .. I will write them down on papers because they are treasure

xvxvxv
Sep 28, 2014

#6**+5 **

Good. I think that they are all ok but I really would like Alan to check that it is all correct.

Anyway, good luck with your test.

Melody
Sep 28, 2014

#7**0 **

in d .. the accelearation is positive and decreasing

in f .. the acceleration is negative and increasing

right .?

xvxvxv
Sep 29, 2014

#8**0 **

the answer in the model answer is b

but the acceleration not constant here because the object covered more ground in each equal time period like the picture above

xvxvxv
Sep 29, 2014

#9**+5 **

I think in d of my little pics the acceleration is negative because the slope of the tangents are negative.

and decreasing (getting closer to 0, I suppose that is really increasing, how confusing) because the gradent of the tangents is becoming less steep with time, the curve is concave up.

------------

In f the acceleration is also negative but the it is increasing - becoming more negative because the gradient is getting steeper - the curve is concave down.

Melody
Sep 29, 2014

#10**0 **

For the middle question the answer is A. The net force on the mass is in the positive x-direction and is constant (assuming static and moving friction are the same). With a constant mass, this means the *acceleration* is constant.

The psychologically misleading aspect of the question is that it is difficult to think in terms of *acceleration*; we are more inclined to see think in terms of *velocity*, so at first imagine the mass slowing down and then speeding up again. It does this, of course, but we now have velocity in mind instead of acceleration. Well, that's my excuse!!

Melody is right on the first and third questions.

Alan
Sep 29, 2014

#11**0 **

Hi Alan,

I didn't understand the middle one partly because I didn't/don't understand what is happening.

I assume the glider is getting pulled up the slope with constant Force.

Since force = mass x acceleration

The mass is constant so the acceleration is constant. That makes sense, at least as far as the formula goes.

So even if the velocity is constant there is still acceleration??

I am afraid this does not make sense to me.

**Could you please attempt to explain this to me Alan?**

-------------------------------

Once the slider is at the top, it is then moved back down again. Was this meant to be considered in this problem. I thought it was and that completely threw me because I don't know what force it returns with. Maybe gravity ??

Maybe I was not suppose to consider the downward motion??

-------------------------------

Melody
Sep 29, 2014

#12**0 **

Alan, I have tried to give you points for your last answer but the system is playing up and it will not allow it.

Sorry.

Melody
Sep 29, 2014

#13**+5 **

The mass is moving up the slope with an initial non-zero velocity. The net-force on the mass is *down* the slope, all the way through the motion.

If velocity is taken as positive in the positive x-direction then for the uphill part of the motion velocity is going from, say, v1=-2m/s at time t1=0s to v2=0 m/s at time t2 = 1s (say). The acceleration is therefore (v2-v1)/(t2-t1) = (0 - (-2))/(1 - 0) = +2m/s^{2}.

We don't intuit acceleration very well!!

Alan
Sep 30, 2014

#14**0 **

Thanks alan, I think that is all a bit too much for me.

Never mind, maybe I will work it out some time in the future.

Melody
Sep 30, 2014

#15**0 **

Alan, for the middle question

1- you made the net force constant because the track is fricitionless ?

2- how did you determine that the acceleration is positive ?

xvxvxv
Oct 8, 2014

#16**+5 **

Best Answer

*1- you made the net force constant because the track is fricitionless ? *

*2- how did you determine that the acceleration is positive ?*

Identify the forces on the glider. Without friction (or air drag etc) the only forces are gravity and the normal reaction force of the track on the glider.

The reaction force balances the component of the gravitational force normal to the track (I'm assuming the glider stays on the track!).

The component of the gravitational force parallel to the track is the *only* force acting to change the speed of the glider, and, since gravity and the slope of the track are both constant, the force is constant. This force acts in the downhill direction which is in the positive direction. The acceleration is therefore in this direction.

**Acceleration doesn't necessarily lead to an increase in speed**. Because the glider starts with an initial speed **up** the track the positive acceleration first acts to slow it down, before bringing it to a halt and **then** increasing its speed back downhill.

.

Alan
Oct 8, 2014