$textbf{Newton’s second law}$ states that the force $F$ on an object is equal to the mass $m$ of the object times its acceleration $a$.

According to this, if a nonzero force acts on an object (mass $m$), it must have a nonzero acceleration, $a = F/m$ ($F$ is nonzero and $m$ can not be $zero$).

According to $textbf{Newton’s second law}$, if a nonzero force acts on an object it must have a nonzero acceleration.

From $textbf{Newton’s second law}$, the force $F$ acting on an object of mass $m$ moving with an acceleration $a$ is given by:

$$
begin{gather*}
F = matag{1}
end{gather*}
$$

The position-time equation for constant acceleration is expressed as:

$$
begin{gather*}
x_f = x_i + v_i t + dfrac{1}{2}at^2tag{2}
end{gather*}
$$

The mass of the shopping cart is $m = 12.3$ kg and the force you exert on it is $F = 10.1$ N.

Its initial speed is $v_i = 0$ and the time of the motion is $t = 2.50$ s.

(a) Rearranging equation (1), we get:

$$
begin{gather*}
a = dfrac{F}{m}
end{gather*}
$$

Now, we plug our values for $m$ and $F$, so we get the acceleration of the cart:

$$
begin{gather*}
a = dfrac{10.1text{ N}}{12.3text{ kg}} = 0.821text{ m/s}^2\\
therefore quad large boxed{a = 0.821text{ m/s}^2}
end{gather*}
$$

So, we plug our values for $v_i$, $a$ and $t$ into equation (2), so we get the distance the cart travels during this interval:

$$
begin{align*}
d & = x_f – x_i\
& = v_i t + dfrac{1}{2}at^2\
& = 0 + dfrac{1}{2} (0.821text{ m/s}^2) (2.50text{ s})^2\
& = 2.57text{ m}
end{align*}
$$

$$
therefore quad large boxed{d = 2.57text{ m}}
$$

(b) $d = 2.57$ m