Verified Solution • 2025–26 Session moderate

QJEEM22Ph2681: JEE Main 2022 Physics - Class 11 JEEM22Ph2681 Solution: Two buses P and Q start from a point at the same time and move…

Two buses P and Q start from a point at the same time and move in a straight line and their positions are represented by XP(t)=αt+βt2{X_P}(t) = \alpha t + \beta {t^2} and XQ(t)=ft−t2{X_Q}(t) = ft - {t^2} . At what time, both the buses have same velocity?

Multiple Choice Options

  • A α−f1+β{{\alpha - f} \over {1 + \beta }}
  • B α+f2(β−1){{\alpha + f} \over {2(\beta - 1)}}
  • C α+f2(1+β){{\alpha + f} \over {2(1 + \beta )}}
  • D f−α2(1+β){{f - \alpha } \over {2(1 + \beta )}}

Step-by-Step Verified Solution

XP=αt+βt2{X_P} = \alpha t + \beta {t^2}

XQ=ft−t2{X_Q} = ft - {t^2}

∴\therefore VP=α+2βt{V_P} = \alpha + 2\beta t

VQ=f−2t{V_Q} = f - 2t

∵\because VP=VQ{V_P} = {V_Q}

⇒α+2βt=f−2t\Rightarrow \alpha + 2\beta t = f - 2t

⇒t=f−α2(1+β)\Rightarrow t = {{f - \alpha } \over {2(1 + \beta )}}

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