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**Hint:**Since velocity is a vector, the change in direction, and not only change in the magnitude, also affects the acceleration of the body. We need to make the x axis parallel to the east and the y axis parallel to the north, and represent the velocities in terms of unit vectors \[\hat i\] and \[\hat j\].

Formula used: In this solution we will be using the following formulae;

\[\left| V \right| = \sqrt {V_x^2 + V_y^2} \] where \[\left| V \right|\] is the magnitude of a vector, whose x and y components are \[{V_x}\] and \[{V_y}\] respectively.

\[\Delta v = {v_2} - {v_1}\] where \[\Delta v\] signifies change in velocity and \[{v_2}\], \[{v_1}\] are the final and initial velocity respectively.

\[{a_v} = \dfrac{{\left| {\Delta v} \right|}}{t}\] where \[{a_v}\] is the magnitude of the average acceleration, and \[t\] is time, \[\left| {\Delta v} \right|\] signifies magnitude of the change in velocity.

**Complete Step-by-Step solution:**

We have a body which changes velocity from 30 m/s due east to 40 m/s due north. To calculate the change in velocity, let’s pick a form of coordinate system in which the x axis is parallel to the direction of the east, and thus y axis is parallel to north. This way, the velocity in the east direction can be written as

\[{v_1} = 30\hat i\] and similarly the due velocity due north can be written as

\[{v_2} = 40\hat j\]

Hence, the change in velocity which is given by

\[\Delta v = {v_2} - {v_1}\] will hence be written as

\[\Delta v = 40\hat j - 30\hat i\]

The magnitude of any vector can be given as

\[\left| V \right| = \sqrt {V_x^2 + V_y^2} \] where \[\left| V \right|\] is the magnitude of a vector, whose x and y components are \[{V_x}\] and \[{V_y}\] respectively.

Hence, for the change in velocity, we have

\[\left| {\Delta v} \right| = \sqrt {{{40}^2} + {{30}^2}} = 50\]

The average acceleration (in magnitude can be given as

\[{a_v} = \dfrac{{\left| {\Delta v} \right|}}{t}\] where \[{a_v}\] is the magnitude of the average acceleration, and \[t\] is time,

Hence,

\[{a_v} = \dfrac{{50}}{{10}} = 5m/{s^2}\]

**Note:**Alternatively, the magnitude of the average acceleration can be calculated by first finding the acceleration in vector, then finding the magnitude, as follows

\[a = \dfrac{{\Delta v}}{t} = \dfrac{{40\hat j - 30\hat i}}{{10}} = 4\hat j - 3\hat i\]

\[{a_v} = \sqrt {{4^2} + {3^2}} = 5m/{s^2}\]

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