P01-060 – Average Acceleration Vector

Average Acceleration Vector

Average acceleration vector for one-dimensional motion:

While the average velocity vector describes how fast the position vector changes over time, the average acceleration vector describes how fast the velocity vector changes over time.

Consider a particle with velocity ${\overrightarrow{v}}_i=v_{ix}\ \hat{x}$ at time $t_i$ and a velocity ${\overrightarrow{v}}_f=v_{fx}\ \hat{x}$ at time $t_f$.

The average acceleration ${\overrightarrow{a}}_{avg\ x}$ of an object during the time interval $\mathrm{\Delta }t=t_f-t_i$ over which its velocity changes by an amount $\mathrm{\Delta }\overrightarrow{v}={\overrightarrow{v}}_f-{\overrightarrow{v}}_i$, is the vector defined by

\boxed{{\overrightarrow{a}}{avg\ x}=\frac{\mathrm{\Delta }\overrightarrow{v}}{\mathrm{\Delta }t}=\frac{{\overrightarrow{v}}_f-{\overrightarrow{v}}_i}{t_f-t_i}=\frac{v_{fx}-v_{ix}}{t_f-t_i}\ \hat{x}\ \ \ \ \ \ \ \ (m/s^2)}

The average acceleration represents how fast the velocity of the particle changes and its sign indicates the direction of the change.

If the velocity of the particle increases from $v_{ix}$ to $v_{fx}>v_{ix}$ in time $\mathrm{\Delta }t=t_f-t_i$ then its average acceleration is positive and equal to

a_{avg\ x}=\frac{v_{fx}-v_{ix}}{t_f-t_i}\ > 0

For example, in the figure below, the particle’s velocity increases from $2\ m/s$ to $4\ m/s$ in time $\mathrm{\Delta }t=2\ s$ and therefore it has an acceleration $a_{avg\ x}=2\ m/s^2$.

Note: an average acceleration of $2\ m/s^2$ means that, on average, the velocity of the particle increases by $2\ m/s$ every second.

If the velocity of the particle decreases from $v_{ix}$ to $v_{fx}<v_{ix}$ in time $\mathrm{\Delta }t=t_f-t_i$ then its average acceleration is negative and equal to

a_{avg\ x}=\frac{v_{fx}-v_{ix}}{t_f-t_i} < 0

For example, in the figure below, the particle’s velocity decreases from $4\ m/s$ to $1\ m/s$ in time $\mathrm{\Delta }t=1\ s$ and therefore it has an acceleration $a_{avg\ x}=-3\ m/s^2$

Note: an average acceleration of $-3\ m/s^2$ means that, on average, the velocity of the particle decreases by $3\ m/s$ every second.

In both instances, the average acceleration is written $\displaystyle{a_{avg\ x}=\frac{v_{fx}-v_{ix}}{t_f-t_i}}$. However, the sign of $a_{avg\ x}$ indicates the direction in which the change in velocity takes place over the time $\mathrm{\Delta }t=t_f-t_i$.