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# 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

Last updated date: 29th Feb 2024
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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
$a = \dfrac{{\Delta v}}{t} = \dfrac{{40\hat j - 30\hat i}}{{10}} = 4\hat j - 3\hat i$
${a_v} = \sqrt {{4^2} + {3^2}} = 5m/{s^2}$