Question

A body is moving with velocity 30m/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: Average acceleration is defined as the difference in the final and initial velocities of the body upon the time taken by the body to reach the final velocity from its initial velocity. Velocity is a vector quantity thus acceleration is also a vector quantity.

Formula used: $\overrightarrow{a}=\dfrac{\overrightarrow{v}-\overrightarrow{u}}{t}$
$R=\sqrt{{{A}^{2}}+{{B}^{2}}+2AB\cos \theta }$

Complete step by step answer:
Acceleration is the change in velocity in one unit of time. It tells us the rate at which the velocity of a given body is increasing or decreasing with respect to time. In other words, how much the velocity is increasing or decreasing in one unit time.
Average acceleration is defined as the difference in the final and initial velocities of the body upon the time taken by the body to reach the final velocity from its initial velocity.
i.e. $a=\dfrac{v-u}{t}$ ……(i),
where a is the acceleration of the body, u and v are initial and final velocities respectively and t is the time taken by the body.
Since, velocity and acceleration are vectors quantities. Equation (i) is written as,
$\overrightarrow{a}=\dfrac{\overrightarrow{v}-\overrightarrow{u}}{t}$ ……..(ii).
To find the average acceleration ($\overrightarrow{a}$) of the given body, let us first find the value of $\overrightarrow{v}$, $\overrightarrow{u}$ and t.
The initial velocity of the body is given to be 30$m{{s}^{-1}}$ towards east. Therefore, $\overrightarrow{u}=30\widehat{i}\text{ }m{{s}^{-1}}$.
The final velocity of the body is given to be 40$m{{s}^{-1}}$ towards the north. Therefore, $\overrightarrow{v}=40\widehat{j}\text{ }m{{s}^{-1}}$.
The time taken by the body to reach this final velocity is given as 10 seconds. Therefore, t=10s.
Substitute the values of $\overrightarrow{v}$, $\overrightarrow{u}$ and t in equation (ii).
$\Rightarrow \overrightarrow{a}=\dfrac{40\widehat{j}-30\widehat{i}}{10}=\dfrac{40\widehat{j}}{10}-\dfrac{30\widehat{i}}{10}=\left( 4\widehat{j}-3\widehat{i} \right)m{{s}^{-2}}$
Therefore, the average acceleration of the body has two components, one with magnitude of 4$m{{s}^{-2}}$ towards north other with magnitude of 3$m{{s}^{-2}}$ towards west.
The magnitude of the resultant of these two components can be found by using the parallelogram law of vectors i.e. $R=\sqrt{{{A}^{2}}+{{B}^{2}}+2AB\cos \theta }$, where A, B and R are the magnitudes of the two components and the resultant respectively. $\theta $ is the angle between the two components.
Here, $\theta =\dfrac{\pi }{2}$ and $\cos \dfrac{\pi }{2}=0$.
Therefore, $R=\sqrt{{{(4)}^{2}}+{{(3)}^{2}}+0}=\sqrt{25}=5m{{s}^{-2}}$.
Therefore, the magnitude of the average acceleration of the body is $5m{{s}^{-2}}$.

The angle that this vector makes with the x-axis is $\alpha $ and from the figure we know that $\tan \alpha =\dfrac{A}{B}=\dfrac{4}{3}$.
$\Rightarrow \alpha ={{\tan }^{-1}}\left( \dfrac{4}{3} \right)$ .

Note: If we do not want to use vector notations, we can directly draw the vectors of velocity and work to find the resultant of the average acceleration. Angle we can find by the same method i.e. from trigonometric ratio tan.