Question

If a particle is moving in such a way that it's average acceleration turns out to be different for a number of different time intervals, the particle is said to have variable acceleration. The acceleration can vary in magnitude, or in direction or both. In such cases we find acceleration at any instant, called the instantaneous acceleration. It is defined as $\overrightarrow{a}=\underset{\Delta t=0}{\text{lim}}\frac{\Delta \overrightarrow{v}}{\Delta t}=\frac{d\overrightarrow{v}}{dt}$
That is acceleration of a particle at time $t$ is the limiting value of $\frac{\Delta v}{\Delta t}$at time $t$ as $\Delta t$ approaches zero. The direction of the instantaneous acceleration $\overrightarrow{a}$ is the limiting direction of the vector in velocity $\Delta v$.

A particle is travelling due north with $5m{s}^{-1}$. it turns east in $5$ s. and continues to move with $5m{s}^{-1}$. Find the average acceleration during turning is

• A. $0$
• B. $\sqrt{2}m{s}^{-2}NE$
• C. $\sqrt{2}m{s}^{-2}SE$
• D. $2m{s}^{-2}SE$

Answer 1

The correct option is C $\sqrt{2}m{s}^{-2}SE$

let $\Delta \overrightarrow{v}=5\hat{i}-5\hat{j},\Delta t=5sec$
The acceleration, $a=\frac{\Delta \overrightarrow{v}}{\Delta t}$
$|a|=\frac{\sqrt{5^2+5^2}}{5}=\sqrt{2}m{s}^{-2}$
The direction of acceleration is the direction of the resultant motion,i.e SE.
Ans:(C)