will u consider 0 as constant acceleration?
#velocity time graph
gray pond
fair parcelBOT
will u consider 0 as constant acceleration?
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gray pond
1- false, v=0
2- false , v=4 at the last sec, before that , its increasing at a constant rate
- false, acceleration means movement, ans its 0 , no movement, no acceleration, theres nothing , no motion happening, rest. ik im wrong so please correct me
- true, negative slope
gray pond
kind leaf
Ok, so, here’s a breakdown:
- You’re right.
- Right answer, wrong reasoning. Think of it like an open interval rather than a closed one.
- Did you calculate the average acceleration for the time period? Is it the same for the entire period?
- You’re right.
@gray pond
kind leaf
The slope is the average velocity.
gray pond
Ok, so, here’s a breakdown:
1. You’re right.
2. Right answer, wrong reasoning. T...
what does open interval and closed interval mean
kind leaf
what does open interval and closed interval mean
Without and with endpoints, respectively.
gray pond
Ok, so, here’s a breakdown:
1. You’re right.
2. Right answer, wrong reasoning. T...
- no its not same for all other time intervals,for 0-3sec a=4/3m/s^2 ( a= (4-0)/3 = 4/3 ) , 3-8sec a=0, 8-12sec a= (0-4)/4 = -1m/s^2
gray pond
The slope is the average velocity.
then acceleration is also the same?
kind leaf
then acceleration is also the same?
What do you mean?
gray pond
The slope is the average velocity.
the slope of v-t graph gives the value of acceleration
so from that statement i assumed u meant to say that slope gives average velocity
velocity only reaches 4 m/s at t = 3 s — it is not 4 m/s throughout, thats what i thought, but now im confused why its 4m/s throughout, acc to Tommy im learning smt new, which is confusing,
what is happening to the value of v as we move ahead ,every millisecond,
he says its same, i.e. 4m/s
he sent me this big ans:
Is v = 4 m/s for all times between t = 0 to 3 s? — YES
The graph shows a horizontal line from t = 0 to t = 3.
A horizontal line on a velocity-time graph means velocity is constant.
And it's sitting at v = 4 m/s the entire time.
So, at t = 0 s, t = 1 s, t = 2.999 s → the velocity is always 4 m/s.
✔️ That's why the answer is Yes.
❌ 2. Is every point on the y-axis = 4 m/s during 0–3 s? — NO
Let’s be careful here.
The y-axis represents velocity values (not time).
Saying “every point on the y-axis = 4 m/s” is like saying:
"All velocities from 0 to 4 m/s happened between t = 0 to 3 s."
That’s not true — the object stayed only at 4 m/s, not 3, not 2, etc.
So the only point on the y-axis that applies in the 0–3 s range is:
v = 4 m/s, and nothing else.
✔️ That’s why the answer is No — only v = 4 matters here.
❌ 3. Does v = 3 m/s at any point between 0–3 s? — NO
Between t = 0 and t = 3, the velocity is locked at 4 m/s (flat line).
There's no change in velocity, so it never passes through 3 m/s.
To have v = 3 m/s, the line would need to be sloped (showing changing velocity), which it isn't.
✔️ So the answer is No — v = 3 m/s doesn’t happen in that time.
from gpt, which i dont understand
kind leaf
from gpt, which i dont understand
https://discord.com/channels/624314920158232616/1378764024094920856
Do not use or analyze the use of LLMs within help threads, as this leads to off-topic discussion and the potential incorrect analysis of questions.
kind leaf
Ok, so, here’s a breakdown:
1. You’re right.
2. Right answer, wrong reasoning. T...
I think the only thing you got wrong is 3., everything else you did right
kind leaf
3. no its not same for all other time intervals,for 0-3sec a=4/3m/s^2 ( a= (4-0)...
But it is asking for only 3 to 8 seconds.
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@gray pond