anonymous
  • anonymous
Kinematics -graphs
Physics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
Acceleration-time graph |dw:1342512372407:dw| Velocity time graph |dw:1342512427429:dw| Now about the position time graph
anonymous
  • anonymous
|dw:1342512503350:dw| which curves(of the 3) will be steeper?
anonymous
  • 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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anonymous
  • anonymous
ok, so u want to know how will the position-time graph look like?
anonymous
  • anonymous
yeah
anonymous
  • anonymous
I'm not sure, but maybe it looks like this: |dw:1342513483775:dw| correct me if i'm wrong.
anonymous
  • anonymous
|dw:1342513648649:dw| ace remains positive always
anonymous
  • 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
  • anonymous
the graph should be steeper from origin to t1 than in the interval t1 - t2.
anonymous
  • anonymous
*t1 to t2
anonymous
  • anonymous
@Ishaan94 the velocity is NOT constant from t1 to t2. accel is.
anonymous
  • anonymous
velocity is NEVER constant
anonymous
  • anonymous
|dw:1342513895634:dw| The graph should look somewhat like this I want to know the order of their steepness.
anonymous
  • anonymous
i think the graph should be of this form: |dw:1342513986456:dw|
anonymous
  • anonymous
1 will be more steep than 2.
anonymous
  • anonymous
the ace is always positive so the velocity always increases and so the slope of xt graph always increases
anonymous
  • anonymous
sry my bad! i saw the slop of accel!
anonymous
  • anonymous
ace decreases t2 onward but stays positive notice
anonymous
  • 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
  • anonymous
the rate of increase of velocity is max from t1 to t2 because ace is max(constant)
anonymous
  • anonymous
and from t2 the velocity increases but at a decreasing rate
anonymous
  • anonymous
@Ishaan94 you lost me
anonymous
  • anonymous
even after t2 ace is positive so slope cant be curving down right?
anonymous
  • 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
  • anonymous
but if ace is positive xt curve never curves down
anonymous
  • 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
  • anonymous
|dw:1342514919696:dw|
anonymous
  • anonymous
but that works only for constant ace part what about the rest?
anonymous
  • anonymous
and i just want the order @Ishaan94
anonymous
  • anonymous
@Ishaan94 yea, i think we can leave the precision out for this sum. just an approximation ill suffice.
anonymous
  • anonymous
okay, here's my final guess: 2 > 1 > 3.
anonymous
  • anonymous
or should it be 2 > 1 = 3 ?
anonymous
  • anonymous
guess?
anonymous
  • anonymous
*answer!
anonymous
  • anonymous
Lets keep the question open for more opinions
anonymous
  • 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
  • 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
anonymous
  • anonymous
Oops wrong curve, from t1 to t2, it should be |dw:1342517034897:dw|A parabola.
anonymous
  • anonymous
so the answer is 2 > 1 > 3 right?
anonymous
  • 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
  • anonymous
Darwin... this is so embarrassing :(
anonymous
  • anonymous
it tends to go downwards. but doesn't do so completely.
anonymous
  • anonymous
anonymous
  • anonymous
so my graph from t2 isn't right either
anonymous
  • anonymous
i hope my math is right
anonymous
  • anonymous
i go with @Ishaan94
anonymous
  • anonymous
mukushla just to be sure of my answer. can you draw an approximate graph? please.
Vincent-Lyon.Fr
  • 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
  • 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
  • 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
  • anonymous
And it's decreasing from \(t_2\). Correct me if I am wrong. :|
anonymous
  • anonymous
Oh silly me :(
anonymous
  • anonymous
Please Ignore me :(
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
@Ishaan what you say is true of velocity, but wrong about displacement
anonymous
  • anonymous
I know :/ I got confused :/
anonymous
  • 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
  • anonymous
@Vincent-Lyon.Fr
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
  • anonymous
Thanks

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