How would zero velocity and nonzero acceleration be represented on a velocity-time graph?

My physics teacher did mention that with zero velocity and non-zero acceleration, the resulting motion would be "speeding up". I don't quite understand this comment as well as how to graph such motion on a velocity-time graph.
Thanks!

1 Answer
Dec 30, 2016

Answer:

Zero velocity and nonzero acceleration are two different cases.

Explanation:

  1. An object having zero velocity implies that it is stationary at a particular location. If we plot it on a velocity-time graph which represents velocity on the #y#-axis and time on the #x#-axis, it is represented by #x#-axis
    Value of #y# (velocity), is #0# for all values of #x# (time).
    graph{y=0x [-2, 27.49, -2, 9.49]}
    Represented by blue line in the figure above.
  2. Nonzero acceleration, assuming it is constant. It could be either positive or negative. The general kinematic equation representing such a motion is

    #v=u+at#
    #v# is final velocity after time #t#, #a# is the acceleration and #u# is the initial velocity.

Compare it with the equation of a straight line in the slope-intercept form

#y=mx+c#
We see that slope of the line represents acceleration.

http://2.bp.blogspot.com
As shown in the figure above, positive slope represents acceleration, (speeding up) from #t=0# to #t=15s#.
Slope is negative from #t=15# to #t=25s#. This represents deceleration or retardation (slowing down).