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shayanreloaded Group TitleBest ResponseYou've already chosen the best response.0
Accelerationtime graph dw:1342512372407:dw Velocity time graph dw:1342512427429:dw Now about the position time graph
 2 years ago

shayanreloaded Group TitleBest ResponseYou've already chosen the best response.0
dw:1342512503350:dw which curves(of the 3) will be steeper?
 2 years ago

shayanreloaded Group TitleBest ResponseYou've already chosen the best response.0
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
 2 years ago

Vaidehi09 Group TitleBest ResponseYou've already chosen the best response.1
ok, so u want to know how will the positiontime graph look like?
 2 years ago

shayanreloaded Group TitleBest ResponseYou've already chosen the best response.0
yeah
 2 years ago

Vaidehi09 Group TitleBest ResponseYou've already chosen the best response.1
I'm not sure, but maybe it looks like this: dw:1342513483775:dw correct me if i'm wrong.
 2 years ago

shayanreloaded Group TitleBest ResponseYou've already chosen the best response.0
dw:1342513648649:dw ace remains positive always
 2 years ago

Vaidehi09 Group TitleBest ResponseYou've already chosen the best response.1
yea, i was going to type in that right now. the max point should not be reached. since V is not 0 anywhere.
 2 years ago

Vaidehi09 Group TitleBest ResponseYou've already chosen the best response.1
the graph should be steeper from origin to t1 than in the interval t1  t2.
 2 years ago

Vaidehi09 Group TitleBest ResponseYou've already chosen the best response.1
*t1 to t2
 2 years ago

Vaidehi09 Group TitleBest ResponseYou've already chosen the best response.1
@Ishaan94 the velocity is NOT constant from t1 to t2. accel is.
 2 years ago

shayanreloaded Group TitleBest ResponseYou've already chosen the best response.0
velocity is NEVER constant
 2 years ago

shayanreloaded Group TitleBest ResponseYou've already chosen the best response.0
dw:1342513895634:dw The graph should look somewhat like this I want to know the order of their steepness.
 2 years ago

Vaidehi09 Group TitleBest ResponseYou've already chosen the best response.1
i think the graph should be of this form: dw:1342513986456:dw
 2 years ago

Vaidehi09 Group TitleBest ResponseYou've already chosen the best response.1
1 will be more steep than 2.
 2 years ago

shayanreloaded Group TitleBest ResponseYou've already chosen the best response.0
the ace is always positive so the velocity always increases and so the slope of xt graph always increases
 2 years ago

Vaidehi09 Group TitleBest ResponseYou've already chosen the best response.1
sry my bad! i saw the slop of accel!
 2 years ago

shayanreloaded Group TitleBest ResponseYou've already chosen the best response.0
ace decreases t2 onward but stays positive notice
 2 years ago

Vaidehi09 Group TitleBest ResponseYou've already chosen the best response.1
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.
 2 years ago

shayanreloaded Group TitleBest ResponseYou've already chosen the best response.0
the rate of increase of velocity is max from t1 to t2 because ace is max(constant)
 2 years ago

shayanreloaded Group TitleBest ResponseYou've already chosen the best response.0
and from t2 the velocity increases but at a decreasing rate
 2 years ago

shayanreloaded Group TitleBest ResponseYou've already chosen the best response.0
@Ishaan94 you lost me
 2 years ago

shayanreloaded Group TitleBest ResponseYou've already chosen the best response.0
even after t2 ace is positive so slope cant be curving down right?
 2 years ago

Vaidehi09 Group TitleBest ResponseYou've already chosen the best response.1
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.
 2 years ago

shayanreloaded Group TitleBest ResponseYou've already chosen the best response.0
but if ace is positive xt curve never curves down
 2 years ago

Vaidehi09 Group TitleBest ResponseYou've already chosen the best response.1
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.
 2 years ago

shayanreloaded Group TitleBest ResponseYou've already chosen the best response.0
dw:1342514919696:dw
 2 years ago

shayanreloaded Group TitleBest ResponseYou've already chosen the best response.0
but that works only for constant ace part what about the rest?
 2 years ago

shayanreloaded Group TitleBest ResponseYou've already chosen the best response.0
and i just want the order @Ishaan94
 2 years ago

Vaidehi09 Group TitleBest ResponseYou've already chosen the best response.1
@Ishaan94 yea, i think we can leave the precision out for this sum. just an approximation ill suffice.
 2 years ago

Vaidehi09 Group TitleBest ResponseYou've already chosen the best response.1
okay, here's my final guess: 2 > 1 > 3.
 2 years ago

Vaidehi09 Group TitleBest ResponseYou've already chosen the best response.1
or should it be 2 > 1 = 3 ?
 2 years ago

shayanreloaded Group TitleBest ResponseYou've already chosen the best response.0
guess?
 2 years ago

shayanreloaded Group TitleBest ResponseYou've already chosen the best response.0
Lets keep the question open for more opinions
 2 years ago

Vaidehi09 Group TitleBest ResponseYou've already chosen the best response.1
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?
 2 years ago

Ishaan94 Group TitleBest ResponseYou've already chosen the best response.1
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
 2 years ago

Ishaan94 Group TitleBest ResponseYou've already chosen the best response.1
Oops wrong curve, from t1 to t2, it should be dw:1342517034897:dwA parabola.
 2 years ago

Vaidehi09 Group TitleBest ResponseYou've already chosen the best response.1
so the answer is 2 > 1 > 3 right?
 2 years ago

Ishaan94 Group TitleBest ResponseYou've already chosen the best response.1
I am sorry. I can't be sure from what it seems my graph from t2 can't be right. VelocityTime graph form t2 is downward parabola right? a = dv/dt?
 2 years ago

Ishaan94 Group TitleBest ResponseYou've already chosen the best response.1
Darwin... this is so embarrassing :(
 2 years ago

Vaidehi09 Group TitleBest ResponseYou've already chosen the best response.1
it tends to go downwards. but doesn't do so completely.
 2 years ago

Ishaan94 Group TitleBest ResponseYou've already chosen the best response.1
so my graph from t2 isn't right either
 2 years ago

Ishaan94 Group TitleBest ResponseYou've already chosen the best response.1
i hope my math is right
 2 years ago

mukushla Group TitleBest ResponseYou've already chosen the best response.0
i go with @Ishaan94
 2 years ago

Ishaan94 Group TitleBest ResponseYou've already chosen the best response.1
mukushla just to be sure of my answer. can you draw an approximate graph? please.
 2 years ago

VincentLyon.Fr Group TitleBest ResponseYou've already chosen the best response.3
Velocitytime is: dw:1342520614050:dw with 2 halfparabolas between 0 and t1 and between 0 t2 and tfinal
 2 years ago

VincentLyon.Fr Group TitleBest ResponseYou've already chosen the best response.3
Displacementtime 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
 2 years ago

Ishaan94 Group TitleBest ResponseYou've already chosen the best response.1
Seems like I made a fool of myself. But Vincent, slope isn't always increasing. It's constant from \(t_1\) to \(t_2\).
 2 years ago

Ishaan94 Group TitleBest ResponseYou've already chosen the best response.1
And it's decreasing from \(t_2\). Correct me if I am wrong. :
 2 years ago

Ishaan94 Group TitleBest ResponseYou've already chosen the best response.1
Oh silly me :(
 2 years ago

Ishaan94 Group TitleBest ResponseYou've already chosen the best response.1
Please Ignore me :(
 2 years ago

VincentLyon.Fr Group TitleBest ResponseYou've already chosen the best response.3
@Ishaan what you say is true of velocity, but wrong about displacement
 2 years ago

Ishaan94 Group TitleBest ResponseYou've already chosen the best response.1
I know :/ I got confused :/
 2 years ago

shayanreloaded Group TitleBest ResponseYou've already chosen the best response.0
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??
 2 years ago

shayanreloaded Group TitleBest ResponseYou've already chosen the best response.0
@VincentLyon.Fr
 2 years ago

VincentLyon.Fr Group TitleBest ResponseYou've already chosen the best response.3
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
 2 years ago

shayanreloaded Group TitleBest ResponseYou've already chosen the best response.0
Thanks
 2 years ago
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