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.
$$\mathrm{\Delta }v=v_2-v_1$$ where $$\mathrm{\Delta }v$$ signifies change in velocity and $$v_2$$, $$v_1$$ are the final and initial velocity respectively.
$$a_v={\displaystyle \frac{\left|\mathrm{\Delta }v\right|}{t}}$$ where $$a_v$$ is the magnitude of the average acceleration, and $$t$$ is time, $$\left|\mathrm{\Delta }v\right|$$ 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
$$\mathrm{\Delta }v=v_2-v_1$$ will hence be written as
$$\mathrm{\Delta }v=40\hat{j}-30\hat{i}$$
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
$$\left|\mathrm{\Delta }v\right|=\sqrt{{40}^2+{30}^2}=50$$
The average acceleration (in magnitude can be given as
$$a_v={\displaystyle \frac{\left|\mathrm{\Delta }v\right|}{t}}$$ where $$a_v$$ is the magnitude of the average acceleration, and $$t$$ is time,
Hence,
$$a_v={\displaystyle \frac{50}{10}}=5m/s^2$$

Note: Alternatively, the magnitude of the average acceleration can be calculated by first finding the acceleration in vector, then finding the magnitude, as follows
$$a={\displaystyle \frac{\mathrm{\Delta }v}{t}}={\displaystyle \frac{40\hat{j}-30\hat{i}}{10}}=4\hat{j}-3\hat{i}$$
$$a_v=\sqrt{4^2+3^2}=5m/s^2$$