Question

A body is moving with velocity 30 m/s towards east. After 10 seconds its velocity becomes 40 m/s towards the north. The average acceleration of the body is

Answer 1

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
\[a = \dfrac{{\Delta v}}{t} = \dfrac{{40\hat j - 30\hat i}}{{10}} = 4\hat j - 3\hat i\]
\[{a_v} = \sqrt {{4^2} + {3^2}} = 5m/{s^2}\]