The SI unit of acceleration is $m{s}^{-1}$.

In this section, we'll define the term "acceleration," which is frequently used in physics. The acceleration will then be determined using a mathematical equation, and the SI units of acceleration will be established as a result.

Formula Used $\overrightarrow{a}=\frac{\overrightarrow{v_2}-\overrightarrow{v_1}}{t_2-t_1}$

The rate at which the velocity changes in relation to time is referred to as acceleration.
If an object's starting velocity at time $t_1$ and $v_2$ and its ultimate velocity at time $t_2$ and $v_2$ then The object's acceleration is then determined by

$a=\frac{v_2-v_1}{t_2-t_1}\to \left(1\right)$

Any object's acceleration will be a vector quantity since every object's velocity is one as well. In contrast to scalar quantities, which only have magnitude, vector quantities have both direction and magnitude. Any vector quantity must have both magnitude and direction in order to be represented.
The acceleration may be expressed in vector form as

$\overrightarrow{a}=\frac{\overrightarrow{v_2}-\overrightarrow{v_1}}{t_2-t_1}$

The International System of Units (SI) uses seconds as the unit of time and metres per second (m/s) as the measure of velocity. The magnitude of the velocity vector quantity is the only distinction between velocity and speed (i.e., speed is a scalar quantity).
Equation (1)'s formula, which determines the SI unit of acceleration, is as follows:

SI unit of acceleration$=\frac{\mathrm{SI}\text{ unit of velocity }}{\text{ SI unit of time }}=\frac{\left(\frac{\mathrm{m}}{\mathrm{s}}\right)}{\mathrm{s}}=\frac{\mathrm{m}}{\mathrm{s}\times \mathrm{s}}=\frac{\mathrm{m}}{{\mathrm{s}}^2}$

The equation above makes it clear that the SI unit of acceleration is the metre per square second (i.e., $\left.\mathrm{m}/{\mathrm{s}}^2\right)$

Thus, the metre per second (m/s) is not the SI unit for acceleration.
As a result, the assertion is false.

Acceleration is defined as the rate of change in velocity in a particular time.

$acceleration=\frac{velocity}{time}=\frac{m{s}^{-1}}{s}=m{s}^{-2}$

Therefore, the statement is false.