Question

A train is standing on a platform, a man inside a compartment of a train drops a stone. At the same instant train starts to move with constant acceleration. The path of the particle as seen by the person who drops the stone is:

• A. Parabola
• B. Straight line for sometime and parabola for the remaining time
• C. Straight line
• D. Variable path that cannot be defined

Answer 1

The correct option is C Straight line

As at the moment when the stone was dropped the train was stationary so the initial vertical velocity of the stone will be zero even with respect to the observer.

Now with respect to the passenger, the stone will have two acceleration $one$ downward due to $gravity$ and $another$ in opposite motion of train because the passenger is on an $accelerated$ frame of reference.

Now downward displacement is $y=\frac{g{t}^2}{2}$ and horizontal displacement as $x=-\frac{a{t}^2}{2}$ where $a$ is acceleration of train.

We see that both $x$ and $y$ are dependent on $\text{same power of time t i.e.,}{t}^2$ so path will be a straight line. just put $t^2=\frac{2x}{a}$ in exprexxion of $y$ we get path equation as $y=\frac{gx}{a}$ which is a stright line in $xy$ plane

Option C is correct.