Question 24: Motion in 1D

Question

A body starts moving from rest with constant acceleration covers displacement $S_1$ in first $(p-1)$ seconds and $\mathrm{S}_2$ in first $p$ seconds. The displacement $\mathrm{S}_1+\mathrm{S}_2$ will be made in time :

Accepted Answer

Let's denote the constant acceleration with which the body moves as $a$. We know that the displacement $S$ covered by a body starting from rest under a constant acceleration $a$ in time $t$ is given by the equation of motion: $S = \frac{1}{2} a t^2$.
Considering the displacement $\mathrm{S}_1$ in first $(p-1)$ seconds, we apply the equation of motion:
$ \mathrm{S}_1 = \frac{1}{2} a (p-1)^2 $
Similarly, for the displacement $\mathrm{S}_2$ in first $p$ seconds:
$ \mathrm{S}_2 = \frac{1}{2} a p^2 $
To find out the total time it will take to cover the displacement $\mathrm{S}_1+\mathrm{S}_2$, we first find the sum of these two displacements:
$ \mathrm{S}_1+\mathrm{S}_2 = \frac{1}{2} a (p-1)^2 + \frac{1}{2} a p^2 $
Let's simplify this:
$ \mathrm{S}_1+\mathrm{S}_2 = \frac{1}{2} a \left((p-1)^2 + p^2ight) $
$ \mathrm{S}_1+\mathrm{S}_2 = \frac{1}{2} a \left(p^2 - 2p + 1 + p^2ight) $
$ \mathrm{S}_1+\mathrm{S}_2 = \frac{1}{2} a \left(2p^2 - 2p + 1ight) $
$ \mathrm{S}_1+\mathrm{S}_2 = a \left(\frac{2p^2 - 2p + 1}{2}ight) $
If we consider the total displacement $\mathrm{S}_1+\mathrm{S}_2$ is to be covered in a time $t$ seconds from rest, we should set this equal to the equation of motion:
$ \mathrm{S}_1+\mathrm{S}_2 = \frac{1}{2} a t^2 $
Equating the two equations:
$ a \left(\frac{2p^2 - 2p + 1}{2}ight) = \frac{1}{2} a t^2 $
Since $a 
eq 0$, we can simplify by dividing both sides by $ \frac{1}{2} a$:
$ \frac{2p^2 - 2p + 1}{2} = \frac{t^2}{2} $
$ 2p^2 - 2p + 1 = t^2 $
To find $t$, we take the square root of both sides:
$ t = \sqrt{2p^2 - 2p + 1} $
Therefore, the time taken to cover the displacement $\mathrm{S}_1+\mathrm{S}_2$ will be:
$ t = \sqrt{(2p^2 - 2p + 1)}\ s $
Hence, the correct option would be:
Option D: $ \sqrt{(2p^2 - 2p + 1)}\ s $

Answer

$(2 p+1) s$

Answer

$(2 p-1) s$

Answer

$\left(2 p^2-2 p+1\right) s$

Answer

$\sqrt{\left(2 p^2-2 p+1\right)} s$

Motion in 1D

A body starts moving from rest with constant acceleration covers displacement S_1 in first (p-1) seconds and \mathrm{S}_2 in first p seconds. The displacement \mathrm{S}_1+\mathrm{S}_2 will be made in time :

Choose the correct answer: