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Motion in a Straight Line — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsMotion in a Straight Line1 Mark
MCQ
At a metro station, a girl walks up a stationary escalator in time $t_1$. If she remains stationary on the escalator, then the escalator take her up in time $t_2$​​​​​​​. The time taken by her to walk up on the moving escalator will be:
  • A
    $\frac{(\text{t}_1+\text{t}_2)}{2}.$
  • B
    $\frac{\text{t}_1\text{t}_2}{(\text{t}_2-\text{t})}.$
  • ✓
    $\frac{\text{t}_1\text{t}_2}{(\text{t}_2+\text{t}_1)}.$
  • D
    $\text{t}_1-\text{t}_2.$
✓

Answer

Correct option: C.
$\frac{\text{t}_1\text{t}_2}{(\text{t}_2+\text{t}_1)}.$
Let $L$ be the length of the escalator.
Velocity of girl w.r.t. ground $\text{v}_\text{g}=\frac{\text{L}}{\text{t}_1}$
Velocity of escalator w.r.t. ground $\text{v}_\text{e}=\frac{\text{L}}{\text{t}_2}$
Effective Velocity of girl on moving escalator with respect to ground $=\text{v}_\text{g}+\text{v}_\text{e}=\frac{\text{L}}{\text{t}_1}+\frac{\text{L}}{\text{t}_2}=\text{L}\Big[\frac{1}{\text{t}_1}+\frac{1}{\text{t}_2}\Big]$
$\text{v}_\text{ge}=\text{L}\Big[\frac{\text{t}_1+\text{t}_2}{\text{t}_1\text{t}_2}\Big]$
$\therefore$ Time $t$ taken by girl on moving escalator in going up the distance $L$ is
$\text{t}=\frac{\text{distance}}{\text{speed}}=\frac{\text{L}}{\text{L}\Big(\frac{\text{t}_1+\text{t}_2}{\text{t}_1\text{t}_2}\Big)}=\frac{\text{t}_1\text{t}_2}{\text{t}_1+\text{t}_2}$
Hence, verifies the option $(c).$

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