what is the different between the first and the second picture . And what is the answer of the third picture

xvxvxv
Sep 29, 2014

#3**+13 **

Just to confirm Melody's answers:

Q2.1 **A**. Velocity is slope of x-t graph including sign. Slope is most positive at P.

Q2.2 **C**. Speed is magnitude of slope of x-t graph (i.e. ignoring sign). Slope has largest magnitude at R.

Q2.3 **D**. Acceleration is slope of velocity-time graph (including sign). Just before point S the velocity is large and negative; just after point S it is large and positive; therefore the change through S is from large negative to large positive velocities. This change is positive and is larger at S than anywhere else.

Alan
Sep 29, 2014

#1**+13 **

I am sorry I have stuffed the order up!

Anyway the answers for 1 and 3 are below.

The graphs are relevant to all of the questions anyway.

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I always have to think about these very hard. I am sure that there are short cuts.

You have a graph of displacement verses time.

underneath I draw a graph of velocity verses time

and under that I drew accel vs time.

I lined the three up so that P, Q, R, and S are all directly under each other. in the 3 graphs.

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I'm going to consider the velocity graph first.

Now at Q and S the partical is stopped so velocity is 0.

At P the velocity is positive and that looks like a turning point so velocity is maximum.

**P is the only point that has positive velocity**

At R the velocity is negative and again on the displacement graph it is a turning point so it is a minimum.

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Now I go from the velocity graph to the acceleration graph.

At P and R the acceleration is 0

At Q acceleration is negative and a tuning point

That leaves S Acceleration is positive.

**SO S MUST BE THE ANSWER for the greatest acceleration.**

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Lets see if I could have used a short cut.

At S, the displacement is a minimum, so velocity is 0, acceleration is positive. (It is about to go in positive direction so acceleration must be positive.

Maybe you can just work through the logic of each point like this but I would still use the diagrams.

Maybe alan can do it more simply. He can probably just 'see' the answer. LOL

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You already know that I have little confidence in Physics but I believe that is correct. Draw the graphs and see what I mean.

Melody
Sep 29, 2014

#2**+8 **

Q2.2

Well it is stopped at q and s

it has positive velocity at P

It has negative velocity at R

the question asks for the greatest speed. NOT the greatest velocity.

so ignoring the sign is the gradient of the tangent steeper at P or R

it is steeper at R.

**so the greatest speed is at R**

Melody
Sep 29, 2014

#3**+13 **

Best Answer

Just to confirm Melody's answers:

Q2.1 **A**. Velocity is slope of x-t graph including sign. Slope is most positive at P.

Q2.2 **C**. Speed is magnitude of slope of x-t graph (i.e. ignoring sign). Slope has largest magnitude at R.

Q2.3 **D**. Acceleration is slope of velocity-time graph (including sign). Just before point S the velocity is large and negative; just after point S it is large and positive; therefore the change through S is from large negative to large positive velocities. This change is positive and is larger at S than anywhere else.

Alan
Sep 29, 2014

#4**+8 **

Thanks very much Alan.

Alan, I just drew these graphs. I think that they are ok but could you check them please.

Unfortunately the pictures do not seem to be displaying properly tonight. Oh well maybe it will appear properly tomorrow.

Here is an online version. It is the same.

http://gyazo.com/b79726fab45f965ab176ff636a7b3045

thank you

Melody
Sep 29, 2014

#5**+3 **

amazing I can't explain my feeling =D ...

points P and R in the first picture ( x - t ) is a turning point but how did you know that p is the maimum velocity and R is minimum ?

xvxvxv
Sep 29, 2014

#6**0 **

alan I don't understand this sentence

just after point S it is large and positive; therefore the change through S is from large negative to large positive velocities. This change is positive and is larger at S than anywhere else.

xvxvxv
Sep 29, 2014

#7**+10 **

Melody, I think your graphs are mainly ok. I think I might draw the start of the accelerration curve with either a smaller initial value or a zero intial value, and then have a smoother change to a peak before it falls to point P. However, this depends very much on the actual functional form of the x-t curve, which we don't know. It's difficult to be sure from a sketch.

xvxvxv, at S the velocity is zero (see Melody's sketches). Just to the left of S at time t1, say, the velocity is negative, just to the right, at time t2, it is positive. Acceleration is rate of change of velocity, so going from t1 to t2 the acceleration is (positive v - negative v)/(t2-t1). Now, positive - negative is positive, and t2-t1 is positive, so overall the acceleration at S is positive.

Alan
Sep 30, 2014

#8**0 **

Thanks very much Alan,

*** points P and R in the first picture ( x - t ) is a turning point but how did you know that p is the maimum velocity and R is minimum ? ***

I have assumed that P and R are points of inflection.

If this is the case then at P on the x-t graph has the steepest positive gradient hence it will have maximum velocity.

And

R on the x-t graph has the steepest negative gradient hence it will have the greatest negative velocity.

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Remember 3*15,

The gradient on the x-t curve IS dy/dx which IS the velocity.

and

The gradient on the v-t curve is dv/dx which IS the acceleration.

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

Melody
Sep 30, 2014

#9**+5 **

Alan, I am a little confused.

The way I have drawn my v-t graph from the origin to point P is concave up.

If this is correct shouldn't the acceleration (between the origin and P) always be positive and decreasing.

Melody
Sep 30, 2014