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 moving along a straight line with $10m{s}^{-1}$. It takes a U-turn in $5s$ and continues to move along with the same velocity $10m{s}^{-1}$. Find the magnitude of average acceleration during turning.

• A. $0$
• B. $2m{s}^{-2}$
• C. $4m{s}^{-2}$
• D. none of these

Answer 1

The correct option is C $4m{s}^{-2}$

Initial velocity,
$u=10m/s$
Final velocity,
$v=-10m/s$
Thus,
Avg acceleration,
$A_{avg}=\frac{-10-10}{5}=-4m/s^2$

$|A_{avg}|=4m/s^2$