acceleration = O (1 Viewer)

[dolbinau]

whhen was acceleration = 0? I couldn't see any inflexion points

 

[emo-kid-91]

whhen was acceleration = 0? I couldn't see any inflexion points

horizontal point of inflexion at 6,80 - top of the curve. right?

 

[BackCountrySnow]

when dv/dt = 0.

 

[@who]

ay mate. it wasnt the point of inflexion, it was the turning point. i.e. where v^/ =o

 

[dolbinau]

horizontal point of inflexion at 6,80 - top of the curve. right?

No, that was a maximum turning point. And acceleration isn't zero when velocity is 0

 

[Smeegen999]

whhen was acceleration = 0? I couldn't see any inflexion points

it would have been at an inflexion if the graph was of x (displacement). but it was velocity so a=0 at turning points

 

[laurennn91]

yeah it was 80, the maximum.

just imagine if you were drawing the gradient function over the top of it... ie. acceleration

 

[ipod21]

how about the total distanced travelled via simpsons rule?? 1480/3 ??

 

[simone.lee]

if you look at the graph, you'll see that the y-axis is the velocity, which means that the maximum velocity is 80 (at t=6)
which means the acceleration is 0 there because it's the maximum velocity....the graph is in terms of velocity & time, not displacement & time

 

[emo-kid-91]

No, that was a maximum turning point. And acceleration isn't zero when velocity is 0

did you draw the acceleration graph? the curve (or line, really) hit the x-axis, ie. acceleration was zero, at the same point where the maximum turning point was on the velocity graph. at 6.

 

[BackCountrySnow]

what qquestion are we even talking about?

 

[MrKim]

It said when? So im guessing they wanted the time? i wrote down 6secs didnt know if its right or not, meh 1 mark

 

[emo-kid-91]

what qquestion are we even talking about?

6b.

 

[dolbinau]

Damn I see now,oh well 1 mark.