Question

An elevator car whose floor-to-ceiling distance is equal to $2.7m$ starts ascending with a constant acceleration $1.2m{s}^{-2}$. 2.0 s after the start a bolt begins falling from the ceiling of the car. Find the displacement covered by the bolt during the free fall in the reference frame fixed to the elevator shaft.
$\text{A}\text{. 0}\text{.7m}$
$\text{B}\text{. 1}\text{.7m}$
$\text{C}\text{. 2}\text{.7m}$
$\text{D}\text{. 3}\text{.7m}$

Answer 1

Hint: Calculate velocity of elevator car/bolt when bolt just starts falling freely. Calculate time taken by bolt to reach the floor of the elevator car using equations of motion. This time, then, can be used to calculate displacement of bolt with respect to ground/elevator shaft.

Formula used:
$v=u+at$,$s=ut+{\displaystyle \frac{1}{2}}a{t}^2$

Complete step by step answer:
 Let us observe the motion of a bolt from the elevator car. Initially when the elevator car starts moving upward, the bolt also moves upward along with the elevator car. Therefore the velocity of the elevator car and bolt will be the same until the bolt starts falling freely. Hence velocity of bolt when it starts falling is given by:
$v=u+at$
Where
$v=$velocity of the bolt after time $t$
$u=$velocity of bolt at time $t=0$
$a=$uniform acceleration of the bolt
In this question, $u=0m{s}^{-1}$,$a=2m{s}^{-2}$and $t=2.0s$.
Let us substitute the values in the above equation. We get
$v=0+1.2\times 2.0$
$v=2.4m{s}^{-1}$
The bolt will travel the distance from the ceiling of the elevator car to the floor of the elevator car as observed from the elevator car. Therefore it will be displaced by $s=2.7m$as we observe from the elevator car.
We know that
$s=ut+{\displaystyle \frac{1}{2}}a{t}^2$
Where
$s=$displacement of the body in time $t$
$u=$initial velocity of the body
$a=$uniform acceleration of the object
In this question,
$s=-2.7m=$ceiling-to-floor distance of the elevator car(displacement is negative because the bolt travels vertically downward)
$u=2.4-2.4=0m{s}^{-1}=$velocity of the bolt when it starts falling with respect to elevator car
$u$is zero because the elevator car is also moving with the same velocity.
$a=-9.8-1.2=-11.0m{s}^{-2}=$acceleration of bolt with respect to elevator car
$t=$time taken by bolt to reach the floor of the elevator
Substituting these values in the corresponding equation we get
$-2.7=0\times t+{\displaystyle \frac{1}{2}}\times (-11.0)\times t^2$
$\Rightarrow 5.5t^2-2.7=0$
$\Rightarrow t^2=0.49s^2$or $t=0.7s$
Now as we need to calculate displacement with respect to elevator shaft, we substitute the values of $u$, $a$ and $t$ with respect to elevator shaft and calculate displacement using the same formula we just used.
Therefore we have
$u=2.4m{s}^{-1}$, $a=-9.8m{s}^{-2}$ and $t=0.7s$
Substituting these we get
Displacement of bolt with reference to elevator shaft after it starts falling freely,$s^{\prime }=2.4\times 0.7+{\displaystyle \frac{1}{2}}(-9.8)\times 0.49$
$s^{\prime }=-0.72m$
Therefore option A is correct.

Note:
 Time is frame independent quantity i.e. time does not change with frame of reference therefore free fall time for bolt does not change.
When calculating the time of free fall for a bolt, the concept of relativity is used.