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QJEEM24Ph3d20: JEE Main 2024 Physics - Class 11 JEEM24Ph3d20 Solution: A body starts moving from rest with constant acceleration…

A body starts moving from rest with constant acceleration covers displacement S1S_1 in first (p−1)(p-1) seconds and S2\mathrm{S}_2 in first pp seconds. The displacement S1+S2\mathrm{S}_1+\mathrm{S}_2 will be made in time :

Multiple Choice Options

  • A (2p+1)s(2 p+1) s
  • B (2p−1)s(2 p-1) s
  • C (2p2−2p+1)s\left(2 p^2-2 p+1\right) s
  • D (2p2−2p+1)s\sqrt{\left(2 p^2-2 p+1\right)} s

Step-by-Step Verified Solution

Let's denote the constant acceleration with which the body moves as aa . We know that the displacement SS covered by a body starting from rest under a constant acceleration aa in time tt is given by the equation of motion: S=12at2S = \frac{1}{2} a t^2 .

Considering the displacement S1\mathrm{S}_1 in first (p−1)(p-1) seconds, we apply the equation of motion:

S1=12a(p−1)2\mathrm{S}_1 = \frac{1}{2} a (p-1)^2

Similarly, for the displacement S2\mathrm{S}_2 in first pp seconds:

S2=12ap2\mathrm{S}_2 = \frac{1}{2} a p^2

To find out the total time it will take to cover the displacement S1+S2\mathrm{S}_1+\mathrm{S}_2 , we first find the sum of these two displacements:

S1+S2=12a(p−1)2+12ap2\mathrm{S}_1+\mathrm{S}_2 = \frac{1}{2} a (p-1)^2 + \frac{1}{2} a p^2

Let's simplify this:

S1+S2=12a((p−1)2+p2)\mathrm{S}_1+\mathrm{S}_2 = \frac{1}{2} a \left((p-1)^2 + p^2\right) S1+S2=12a(p2−2p+1+p2)\mathrm{S}_1+\mathrm{S}_2 = \frac{1}{2} a \left(p^2 - 2p + 1 + p^2\right) S1+S2=12a(2p2−2p+1)\mathrm{S}_1+\mathrm{S}_2 = \frac{1}{2} a \left(2p^2 - 2p + 1\right) S1+S2=a(2p2−2p+12)\mathrm{S}_1+\mathrm{S}_2 = a \left(\frac{2p^2 - 2p + 1}{2}\right)

If we consider the total displacement S1+S2\mathrm{S}_1+\mathrm{S}_2 is to be covered in a time tt seconds from rest, we should set this equal to the equation of motion:

S1+S2=12at2\mathrm{S}_1+\mathrm{S}_2 = \frac{1}{2} a t^2

Equating the two equations:

a(2p2−2p+12)=12at2a \left(\frac{2p^2 - 2p + 1}{2}\right) = \frac{1}{2} a t^2

Since a≠0a \neq 0 , we can simplify by dividing both sides by 12a\frac{1}{2} a :

2p2−2p+12=t22\frac{2p^2 - 2p + 1}{2} = \frac{t^2}{2} 2p2−2p+1=t22p^2 - 2p + 1 = t^2

To find tt , we take the square root of both sides:

t=2p2−2p+1t = \sqrt{2p^2 - 2p + 1}

Therefore, the time taken to cover the displacement S1+S2\mathrm{S}_1+\mathrm{S}_2 will be:

t=(2p2−2p+1) st = \sqrt{(2p^2 - 2p + 1)}\ s

Hence, the correct option would be:

Option D: (2p2−2p+1) s\sqrt{(2p^2 - 2p + 1)}\ s

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