P01-080 – Instantaneous Acceleration

Instantaneous Acceleration

Instantaneous acceleration for one-dimensional motion:

In the same way we derived the instantaneous velocity from the average velocity vector, we now derive the instantaneous acceleration from the average acceleration vector by letting $\mathrm{\Delta }t\to 0$.

With help from calculus, we define the instantaneous acceleration as the following limit

\boxed{a_x={\mathop{\mathrm{lim}}_{\mathrm{\Delta }t\to 0} \frac{\mathrm{\Delta }v}{\mathrm{\Delta }t}\ }=\frac{dv_x}{dt}\ \ \ \ \ \ \ \ (m/s^2)}

where such a limit is, by definition, equal to the derivative of $v$ with respect to $t$ and is denoted by $\displaystyle{\frac{dv_x}{dt}}$.

• A positive instantaneous acceleration at an instant $t$ denotes an increase in the instantaneous velocity at that instant.

• A negative instantaneous acceleration at an instant $t$ denotes a decrease in the instantaneous velocity at that instant.

Example: if the velocity $v_x$ of a particle as it moves along a given path is described by the function $v_x\left(t\right)=6t-1$, then we can conclude that its instantaneous acceleration is equal to

a_x\left(t\right)=\frac{dv_x}{dt}=6\ \ \ \ \ (m/s^2)

This yields a function of time for $a_x\left(t\right)$ which will output the instantaneous acceleration of the particle at any instant $t$ of its motion. In particular, a constant value of $6\ m/s^2$ indicates that the speed of the particle increases by $6\ m/s$ every second.