This discussion is now closed.
Check out other Related discussions
- SQA Higher Physics - Paper 2 - 25th April 2024 [Exam Chat]
- physics question help please
- A level Mechanics: Variable acceleration differentiation
- WJEC Physics Unit 4 (13/6/2024)
- SQA Higher physics discussion
- Edexcel A Level Physics Paper 1: Advanced Physics I 9PH0 01 - 24 May 2024 [Exam Chat]
- as level mechanics suvat
- is differentiating x in terms of y the same as implicit differentiation?
- Position-time vs distance-time graph
- SQA higher physics prediction
- MAT 1996-2006 Solution Thread
- Using the curves on the force/extension graph transfer to stress/strain graph
- How to do convert a negative velocity to speed? Alevel maths
- Physics edexcel igcse 2024
- Maths question help please
- Quatum Mechanics Primer Isaac Physics
- Stuck on A-level electronics question, please help! :-(
- Can someone direct me to the procedure of this experiment?
- Maths Question
- Mechanics A level question
What does the slope/gradient of an acceleration-time graph indicate?
I understand that the slope of a displacement-time graph would give velocity, as for velocity-time graph it gives acceleration.. and I think it can be proven by the units..
e.g : displacement-time graph
since the unit for displacement is metres, (m) and the unit for time is seconds, (s) ..
m/s = ms^-1 = unit for velocity time
e.g : velocity-time graph
since the unit for velocity is metres per second (m/s) and the unit for time is seconds (s) ..
ms^-1 / s = ms^-2 = unit for acceleration
there fore the slope of a velocity-time graph would give the unit for acceleration.
However I am confused for a acceleration-time graph would give
ms^-2 / s = ms^-3???
what is that?
Have I gotten a totally wrong idea and what does the slope/gradient of an acceleration-time graph indicate then?
e.g : displacement-time graph
since the unit for displacement is metres, (m) and the unit for time is seconds, (s) ..
m/s = ms^-1 = unit for velocity time
e.g : velocity-time graph
since the unit for velocity is metres per second (m/s) and the unit for time is seconds (s) ..
ms^-1 / s = ms^-2 = unit for acceleration
there fore the slope of a velocity-time graph would give the unit for acceleration.
However I am confused for a acceleration-time graph would give
ms^-2 / s = ms^-3???
what is that?
Have I gotten a totally wrong idea and what does the slope/gradient of an acceleration-time graph indicate then?
Reply 1
Yes and No.
Acceleraton is m per second per second and is a 2nd order differential or a differential of a differential.
Try drawing a graph of both f'(x) and f''(x) and you will see why.
the time graph is a second order differential that's all you need to do.
This is the result of an integral.
where
= displacement
= initial velocity
= final velocity
= uniform acceleration
t = time.
edit: I just realised you are doing GCSE maths so this is no help whatsoever. lol.
Acceleraton is m per second per second and is a 2nd order differential or a differential of a differential.
Try drawing a graph of both f'(x) and f''(x) and you will see why.
the time graph is a second order differential that's all you need to do.
This is the result of an integral.
where
= displacement
= initial velocity
= final velocity
= uniform acceleration
t = time.
edit: I just realised you are doing GCSE maths so this is no help whatsoever. lol.
Reply 2
We call it JOLT or sometimes JERK. Usually only significant in rocket technology.
inexplicable_apathy
I understand that the slope of a displacement-time graph would give velocity, as for velocity-time graph it gives acceleration.. and I think it can be proven by the units..
e.g : displacement-time graph
since the unit for displacement is metres, (m) and the unit for time is seconds, (s) ..
m/s = ms^-1 = unit for velocity time
e.g : velocity-time graph
since the unit for velocity is metres per second (m/s) and the unit for time is seconds (s) ..
ms^-1 / s = ms^-2 = unit for acceleration
there fore the slope of a velocity-time graph would give the unit for acceleration.
However I am confused for a acceleration-time graph would give
ms^-2 / s = ms^-3???
what is that?
Have I gotten a totally wrong idea and what does the slope/gradient of an acceleration-time graph indicate then?
e.g : displacement-time graph
since the unit for displacement is metres, (m) and the unit for time is seconds, (s) ..
m/s = ms^-1 = unit for velocity time
e.g : velocity-time graph
since the unit for velocity is metres per second (m/s) and the unit for time is seconds (s) ..
ms^-1 / s = ms^-2 = unit for acceleration
there fore the slope of a velocity-time graph would give the unit for acceleration.
However I am confused for a acceleration-time graph would give
ms^-2 / s = ms^-3???
what is that?
Have I gotten a totally wrong idea and what does the slope/gradient of an acceleration-time graph indicate then?
It's just the rate of change in acceleration, showing that the acceleration isn't constant.
Reply 4
Apologies for that last answer what is ms^-1 per ms^-1?
Or what is m/s/m/s?
Speed over time over time =acceleration
Hence m/s/s=?
Or what is m/s/m/s?
Speed over time over time =acceleration
Hence m/s/s=?
Reply 5
if you're assuming constant mass with acceleration being a=f/m, then an acceleration/time graph can be an analogue of a force/time graph... so yeah, jerk or jolt would be an apt description as an abrupt change in the graph would be a jerk or jolt
Reply 6
Shows that the force being applied changes.. and change in force should be generated by change in work -> power and so on.
Reply 7
It's shows the rate of change of acceleration.
If you think about a = F/m then a varying force will produce this. This happens with a pendulum, or mass on a spring, where the force depends on the displacement of the mass from its equilibrium position. So the acceleration changes with time as the mass oscillates, being greatest at maximum displacement.
You also get non uniform acceleration if the mass changes. Others have mentioned rockets. As fuel is burned, the mass gets smaller. This would cause the acceleration to get larger for the same thrust.
If you think about a = F/m then a varying force will produce this. This happens with a pendulum, or mass on a spring, where the force depends on the displacement of the mass from its equilibrium position. So the acceleration changes with time as the mass oscillates, being greatest at maximum displacement.
You also get non uniform acceleration if the mass changes. Others have mentioned rockets. As fuel is burned, the mass gets smaller. This would cause the acceleration to get larger for the same thrust.
Reply 9
Slope of a n acceleration-time graph gives 'jerk' (jhatka)
Reply 10
This is indicated by slope of an a-t graph:-
It is called a jerk.
It is acceleration/ time
= (M/s²)/s
= (M/s²)×(1/s)
= M/s³(SI UNIT)
Hope this is helpful
It is called a jerk.
It is acceleration/ time
= (M/s²)/s
= (M/s²)×(1/s)
= M/s³(SI UNIT)
Hope this is helpful
(edited 6 years ago)
Related discussions
- SQA Higher Physics - Paper 2 - 25th April 2024 [Exam Chat]
- physics question help please
- A level Mechanics: Variable acceleration differentiation
- WJEC Physics Unit 4 (13/6/2024)
- SQA Higher physics discussion
- Edexcel A Level Physics Paper 1: Advanced Physics I 9PH0 01 - 24 May 2024 [Exam Chat]
- as level mechanics suvat
- is differentiating x in terms of y the same as implicit differentiation?
- Position-time vs distance-time graph
- SQA higher physics prediction
- MAT 1996-2006 Solution Thread
- Using the curves on the force/extension graph transfer to stress/strain graph
- How to do convert a negative velocity to speed? Alevel maths
- Physics edexcel igcse 2024
- Maths question help please
- Quatum Mechanics Primer Isaac Physics
- Stuck on A-level electronics question, please help! :-(
- Can someone direct me to the procedure of this experiment?
- Maths Question
- Mechanics A level question
Latest
Trending
Last reply 3 weeks ago
Isaac Physics question C4a.4 essential pre uni physics. Please help.Last reply 1 month ago
does GPE decrease or increase the closer you get to a planet?
