## shayanreloaded 3 years ago Kinematics -graphs

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

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

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

4. Vaidehi09

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

yeah

6. Vaidehi09

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

|dw:1342513648649:dw| ace remains positive always

8. Vaidehi09

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

9. Vaidehi09

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

10. Vaidehi09

*t1 to t2

11. Vaidehi09

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

velocity is NEVER constant

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

14. Vaidehi09

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

15. Vaidehi09

1 will be more steep than 2.

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

17. Vaidehi09

sry my bad! i saw the slop of accel!

ace decreases t2 onward but stays positive notice

19. Vaidehi09

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.

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

and from t2 the velocity increases but at a decreasing rate

@Ishaan94 you lost me

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

24. Vaidehi09

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.

but if ace is positive xt curve never curves down

26. Vaidehi09

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.

|dw:1342514919696:dw|

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

and i just want the order @Ishaan94

30. Vaidehi09

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

31. Vaidehi09

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

32. Vaidehi09

or should it be 2 > 1 = 3 ?

guess?

34. Vaidehi09

Lets keep the question open for more opinions

36. Vaidehi09

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?

37. Ishaan94

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

38. Ishaan94

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

39. Vaidehi09

so the answer is 2 > 1 > 3 right?

40. Ishaan94

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?

41. Ishaan94

Darwin... this is so embarrassing :(

42. Vaidehi09

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

43. Ishaan94

@mukushla

44. Ishaan94

so my graph from t2 isn't right either

45. Ishaan94

i hope my math is right

46. mukushla

i go with @Ishaan94

47. Ishaan94

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

48. Vincent-Lyon.Fr

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

49. 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

50. Ishaan94

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

51. Ishaan94

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

52. Ishaan94

Oh silly me :(

53. Ishaan94

54. Vincent-Lyon.Fr

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

55. Ishaan94

I know :/ I got confused :/

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??

@Vincent-Lyon.Fr

58. 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