#### Answer

(a) (3) Magnitude of the acceleration vector is between $4\,\mathrm{m/s^2}$ and $7\,\mathrm{m/s^2}$.
(b) The acceleration has a magnitude of $5\,\mathrm{m/s^2}$, directed at angle of $53^{\circ}$ to the x-axis.

#### Work Step by Step

The magnitude of a vector is given by -
$|a|=\sqrt{a_x^2+a_y^2}$
(a) This is analogous to the magnitude being the hypotenuse of a triangle with x- and y-components as the two other sides. Thus, the magnitude of the vector must be greater than the larger of the two components, which is 4 here. And, the magnitude must also be less than the sum of the components, which is 7 here.
(b) Using the values of the components, $a_x=3$ and $a_y=4$ -
$|a|=\sqrt{9+16}=5$.
for direction at an angle $\theta$ to x-axis -
$\tan{\theta}=\frac{a_y}{a_x}=\frac{4}{3}$
or, $\theta=53^{\circ}$.