# 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!

Dec 30, 2016

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

$v = u + a t$
$v$ 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

$y = m x + c$
We see that slope of the line represents acceleration.

As shown in the figure above, positive slope represents acceleration, (speeding up) from $t = 0$ to $t = 15 s$.
Slope is negative from $t = 15$ to $t = 25 s$. This represents deceleration or retardation (slowing down).