Q in instantanous acceleration I want check my answer .

[r-soy]

Hi all

Q in instantaneous acceleration ?



https://www.physicsforums.com/attachments/17036



my answer is :



a ) As we know Instanteouse acceleration :

It is the limit of the average acceleration when the time interval to zero



a = changing in v / changing t
3.0-1.5 / 0.8 - 0.4 = 35225 m/s



b )

instantanous acceleration = Zero because it's straight line



c ) here the gragh go down then the ( a ) will by ( - )



a = changing in v / changing t



3.0 - 1.5 / 1.6 - 1.4 = -0.4625



this is my answer I want check

 

[Dadface]

Using your method part a) comes to 1.5/0.4=3.75m/s^2.Your method is right but something went wrong with your maths(used calculator perhaps and pressed wrong buttons?).Please note that you could have used the whole triangle and written a=3/0.8.

You made a similar mistake in part c)

You need to improve on your reasoning for part b).The acceleration is zero because it is a straight line parallel to the time axis,in other words there is no change of velocity.

 

[r-soy]

hi thanks

but what are the mistake in part c ?

 

[Dadface]

You wrote a=3-1.5/1.6-1.4.It would be clearer if you wrote a=-(3-1.5)/(1.6-1.4)=1.5/0.2=-7.5m/s^2