# What if she steps on the opposite side, walking against the motion of the sidewalk, what must be her speed so that she reaches the end of the track in the same time as in part (a)?

## (a) is 14 s. Xo=0 X= +35 m i A ‘moving sidewalk’ in an airport terminal moves at 1 m/s and is 35 m long. If a woman steps on at one end and walks 1.5 m/s relative to the sidewalk, in the same direction as the movement of sidewalk.

Feb 22, 2017

$3.5 {\text{ ms}}^{-} 1$

#### Explanation:

Let her speed be $v {\text{ ms}}^{-} 1$ as she moves against the motion of sidewalk.
Her relative or effective speed $= \text{Speed" -"Speed of sidewalk}$
$= v - 1 {\text{ ms}}^{-} 1$

Displacement$= 35 - 0 = 35 \text{ m}$

Equating to the time taken in (a)
$14 = \text{Displacement"/"Relative speed}$
$\implies 14 = \frac{35}{v - 1}$
$\implies 14 \times \left(v - 1\right) = 35$

Solving for $v$ we get
$v = \frac{35}{14} + 1 = 3.5 {\text{ ms}}^{-} 1$
.-.-.-.-.-.-.-.-.-.-.-.-.-.-.

Alternate approach

As time and distance moved in both directions is same. Relative speed in both directions must be same.

Relative speed in positive direction, along the direction of motion of sidewalk$= {\text{Forward Speed"+"Speed of sidewalk"=1.5+1=2.5" ms}}^{-} 1$

Relative speed in opposite direction, against the direction of motion of sidewalk$= {\text{Speed"-"Speed of sidewalk"=2.5" ms}}^{-} 1$
$\implies {\text{Speed "v=2.5+1=3.5" ms}}^{-} 1$