Question

A point moving along the x-direction starts from rest at $x=0$ and comes to rest at $x=1$ after $1\text{s}$. Its acceleration at any point is denoted by $\alpha$. Which of the following is not correct?

• A. $\alpha$ must change sign during the motion.
• B. $|\alpha |\ge 4$ units at some or all points during the motion.
• C. It is not possible to specify an upper limit for $|\alpha |$ from the given data.
• D. $|\alpha |$ cannot be less that $\frac{1}{2}$ during the motion.

Answer 1

The correct option is D $|\alpha |$ cannot be less that $\frac{1}{2}$ during the motion.


The simplest solution to the motion is shown by the plot OAB. The area under the $v-t$ plot must be equal to 1 (displacement).
For this, $\frac{v}{2}=1$ or $v=2$
The acceleration $=\frac{v}{\frac{1}{2}}=4\text{units}$.
$\therefore$ Slope of $OA=4$
Taking OAB as reference, the condition of the motion can be satisfied by any other curve, e.g., OCB, as long as the area under it is equal to 1. Also, the slope of the curve at any point gives the acceleration $\alpha$. Points D and E have $\alpha =4$. Points below D have $\alpha <4$ and above D have $\alpha >4$. As the shape of the curve OCB is arbitrary, it is not possible to set upper or lower limits on the acceleration.