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A question about constant velocity/constant acceleration graphs?

  • #1
Hey! For a lab, I have to make a few graphs, I just need some clarification.

Please correct me if I'm wrong.

For a D-T graph, where velocity is constant, the line will be straight. (Diagonal.)
For a V-T graph, where velocity is constant, the line will be straight. (Horizontal)
^ Confused about this though, at point 0 what will the y value be? :eek: (For example, if the velocity is 3m/s)
For a A-T graph, where velocity is constant, there will be NO line. (No acceleration)


For a D-T graph, where acceleration is constant, the line will be curved.
For a V-T graph, where acceleration is constant, the line will be straight. (Diagonal)
For a A-T graph, where acceleration is constant, the line will be straight. (Horizontal)??
^ For this one I am also confused.

I just need to know if I'm on the right track! :blushing:
Ones in bold are the one I'm not sure are correct.

Thank you very much for reading! I appreciate any help that is offered. =)
 

Answers and Replies

  • #2
648
2
Hey! For a lab, I have to make a few graphs, I just need some clarification.

Please correct me if I'm wrong.

For a D-T graph, where velocity is constant, the line will be straight. (Diagonal.)
For a V-T graph, where velocity is constant, the line will be straight. (Horizontal)
^ Confused about this though, at point 0 what will the y value be? :eek: (For example, if the velocity is 3m/s)
The horizontal line will be above the horizontal axis at a distance equal to 3m/s on the scale.
For a A-T graph, where velocity is constant, there will be NO line. (No acceleration)
The line will be horizontal and along the axis where a=0.
For a D-T graph, where acceleration is constant, the line will be curved.
For a V-T graph, where acceleration is constant, the line will be straight. (Diagonal)
For a A-T graph, where acceleration is constant, the line will be straight. (Horizontal)??
^ For this one I am also confused.
The horizontal line will be above the x axis for positive acceleration and below it for negative acceleration.
You are on the right track.

This might help
http://www.bbc.co.uk/scotland/learning/bitesize/higher/physics/mech_matt/analyse_motion_rev1.shtml
 
  • #3
thank you for all your help :)
 
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