Electron Motion
An electron's motion is tracked and plotted in the shown diagram. What is its acceleration between the 2 and 4 second data points in m/s^2?
Expand Hint
Acceleration is defined as:
$$$a=\frac{dv}{dt}$$$
where
$$v$$
is velocity and
$$t$$
is time.
Hint 2
$$$a=\frac{dv}{dt}=\frac{\Delta v}{\Delta t}=\frac{v_{final}-v_{initial}}{t_{final}-t_{initial}}$$$
Acceleration is defined as:
$$$a=\frac{dv}{dt}=\frac{\Delta v}{\Delta t}=\frac{v_{final}-v_{initial}}{t_{final}-t_{initial}}$$$
where
$$v$$
is velocity and
$$t$$
is time.
The plot’s x-axis is time and y-axis is velocity. Looking at the graph to find the corresponding velocity values associated with 2 and 4 second observation points:
Therefore,
$$v_{final}=1\:m/s$$
,
$$v_{initial}=1\:m/s$$
,
$$t_{final}=4\:s$$
, and
$$t_{initial}=2\:s$$
. Plugging these values into the acceleration equation:
$$$a=\frac{1m/s-1m/s}{4s-2s}=\frac{0m/s}{2s}=0\:m/s^2$$$
$$$0\:m/s^2$$$
Time Analysis
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