A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 3 years ago
Kinematics graphs
anonymous
 3 years ago
Kinematics graphs

This Question is Closed

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Accelerationtime graph dw:1342512372407:dw Velocity time graph dw:1342512427429:dw Now about the position time graph

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ace 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok, so u want to know how will the positiontime graph look like?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I'm not sure, but maybe it looks like this: dw:1342513483775:dw correct me if i'm wrong.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1342513648649:dw ace remains positive always

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the graph should be steeper from origin to t1 than in the interval t1  t2.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@Ishaan94 the velocity is NOT constant from t1 to t2. accel is.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0velocity is NEVER constant

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i think the graph should be of this form: dw:1342513986456:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.01 will be more steep than 2.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the ace is always positive so the velocity always increases and so the slope of xt graph always increases

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0sry my bad! i saw the slop of accel!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ace decreases t2 onward but stays positive notice

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i 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
 3 years ago
Best ResponseYou've already chosen the best response.0the rate of increase of velocity is max from t1 to t2 because ace is max(constant)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and from t2 the velocity increases but at a decreasing rate

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@Ishaan94 you lost me

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0even after t2 ace is positive so slope cant be curving down right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i 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
 3 years ago
Best ResponseYou've already chosen the best response.0but if ace is positive xt curve never curves down

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0right 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
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1342514919696:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0but that works only for constant ace part what about the rest?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and i just want the order @Ishaan94

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@Ishaan94 yea, i think we can leave the precision out for this sum. just an approximation ill suffice.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0okay, here's my final guess: 2 > 1 > 3.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0or should it be 2 > 1 = 3 ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Lets keep the question open for more opinions

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0tell 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
 3 years ago
Best ResponseYou've already chosen the best response.0I 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Oops wrong curve, from t1 to t2, it should be dw:1342517034897:dwA parabola.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so the answer is 2 > 1 > 3 right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I 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?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Darwin... this is so embarrassing :(

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0it tends to go downwards. but doesn't do so completely.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so my graph from t2 isn't right either

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i hope my math is right

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0mukushla just to be sure of my answer. can you draw an approximate graph? please.

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Displacementtime 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
 3 years ago
Best ResponseYou've already chosen the best response.0Seems like I made a fool of myself. But Vincent, slope isn't always increasing. It's constant from \(t_1\) to \(t_2\).

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0And it's decreasing from \(t_2\). Correct me if I am wrong. :

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@Ishaan what you say is true of velocity, but wrong about displacement

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I know :/ I got confused :/

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so 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
 3 years ago
Best ResponseYou've already chosen the best response.0The 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
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.