Kinematics -graphs

- anonymous

Kinematics -graphs

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- anonymous

Acceleration-time graph
|dw:1342512372407:dw|
Velocity time graph
|dw:1342512427429:dw|
Now about the position time graph

- anonymous

|dw:1342512503350:dw|
which curves(of the 3) will be steeper?

- anonymous

ace increases from origin to t1 constant from t1 to t2 and decreases from t2 onward.
but ace is max from t1 to t2.
I hope im understandable

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## More answers

- anonymous

ok, so u want to know how will the position-time graph look like?

- anonymous

yeah

- anonymous

I'm not sure, but maybe it looks like this:
|dw:1342513483775:dw|
correct me if i'm wrong.

- anonymous

|dw:1342513648649:dw|
ace remains positive always

- anonymous

yea, i was going to type in that right now. the max point should not be reached. since V is not 0 anywhere.

- anonymous

the graph should be steeper from origin to t1 than in the interval t1 - t2.

- anonymous

*t1 to t2

- anonymous

@Ishaan94 the velocity is NOT constant from t1 to t2. accel is.

- anonymous

velocity is NEVER constant

- anonymous

|dw:1342513895634:dw|
The graph should look somewhat like this
I want to know the order of their steepness.

- anonymous

i think the graph should be of this form:
|dw:1342513986456:dw|

- anonymous

1 will be more steep than 2.

- anonymous

the ace is always positive so the velocity always increases and so the slope of xt graph always increases

- anonymous

sry my bad! i saw the slop of accel!

- anonymous

ace decreases t2 onward but stays positive
notice

- anonymous

i think the order of steepness should be 1 > 2 > 3.
i'm just using logic here:
accel is increasing from 0 to t1 hence vel is increasing at the fastest rate in this interval. from t1 to t2, accel is constant. so vel is increasing at a constant rate ( this rate < the rate from 0 to t1).
and from t2 to the end, accel is decreasing. so the vel is anyway decreasing. so less distance will be covered as compared to the former 2 intervals.

- anonymous

the rate of increase of velocity is max from t1 to t2 because ace is max(constant)

- anonymous

and from t2 the velocity increases but at a decreasing rate

- anonymous

@Ishaan94 you lost me

- anonymous

even after t2 ace is positive so slope cant be curving down right?

- anonymous

i think it should be curving down. because now the distance covered per unit time is decreasing. so 3 should be the least steep of all.

- anonymous

but if ace is positive xt curve never curves down

- anonymous

right now, the interval we're considering is this:
|dw:1342514762867:dw|
so at some point the graph will curve downwards, but we don't have that within our interval.

- anonymous

|dw:1342514919696:dw|

- anonymous

but that works only for constant ace part what about the rest?

- anonymous

and i just want the order @Ishaan94

- anonymous

@Ishaan94 yea, i think we can leave the precision out for this sum. just an approximation ill suffice.

- anonymous

okay, here's my final guess:
2 > 1 > 3.

- anonymous

or should it be 2 > 1 = 3 ?

- anonymous

guess?

- anonymous

*answer!

- anonymous

Lets keep the question open for more opinions

- anonymous

tell me this:
accel is increasing and decreasing at the same rate from 0 to t1 and from t2 to the end. so what happens to the distance covered per unit time during these intervals? will the be equal?

- anonymous

I am sorry for deleting my previous posts. Acceleration is positive from t2 and on wards but I presumed it to be negative :(
My latest attempt. I know it sucks but still

##### 1 Attachment

- anonymous

Oops wrong curve, from t1 to t2, it should be |dw:1342517034897:dw|A parabola.

- anonymous

so the answer is 2 > 1 > 3 right?

- anonymous

I am sorry. I can't be sure from what it seems my graph from t2 can't be right.
Velocity-Time graph form t2 is downward parabola right? a = -dv/dt?

- anonymous

Darwin... this is so embarrassing :(

- anonymous

it tends to go downwards. but doesn't do so completely.

- anonymous

@mukushla

- anonymous

so my graph from t2 isn't right either

- anonymous

i hope my math is right

- anonymous

i go with @Ishaan94

- anonymous

mukushla just to be sure of my answer. can you draw an approximate graph? please.

- Vincent-Lyon.Fr

Velocity-time is:
|dw:1342520614050:dw|
with 2 half-parabolas between 0 and t1 and between 0 t2 and tfinal

- Vincent-Lyon.Fr

Displacement-time is:
|dw:1342520737774:dw|
The curve is always increasing as well as increasing in slope. Only the curvature becomes 0 at the end (would go on in a straight line if acceleration remains 0 or would reach an inflexion point if acceleration becomes negative after tfinal

- anonymous

Seems like I made a fool of myself.
But Vincent, slope isn't always increasing. It's constant from \(t_1\) to \(t_2\).

- anonymous

And it's decreasing from \(t_2\). Correct me if I am wrong. :|

- anonymous

Oh silly me :(

- anonymous

Please Ignore me :(

- Vincent-Lyon.Fr

@Ishaan what you say is true of velocity, but wrong about displacement

- anonymous

I know :/ I got confused :/

- anonymous

so whats the order of steepness??
3>2>1 ? or
1=2<3 ?
does it depend on ace magnitude or rate of change of ace??

- anonymous

@Vincent-Lyon.Fr

- Vincent-Lyon.Fr

The steepness (of displacement curve) is ever increasing because velocity is; so 3>2>1
But, as velocity is a continuous function, the 3 parts of the curve will smoothly
link together, without any angle at t1 and t2

- anonymous

Thanks

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