Question

At an airport, a bored child starts to walk backwards on a moving platform. The child accelerates relative to the platform with $a=-0.5m/s^2$ relative to the platform. The platform moves with a constant speed $v=+1.0m/s$ relative to the stationary floor. In $4.0$ seconds, how much will the child have been displaced relative to the floor?

• A. $8m$
• B. $0m$
• C. $3m$
• D. $4m$

Answer 1

The correct option is B $0m$

initial vel = vel of platform $=1m/s$

acceleration wrt floor $=$ acceleration of him wrt platform +acc. of platform

$=\frac{-1}{2}m/s^2+0=\frac{-1}{2}m/s^2$

So, for finding displacement of him in us,

Applying second equation of motion

$x=ut+\frac{a{t}^2}{2}$

$\left(1\right)(4{)}^2-\left(\frac{1}{4}\right)\frac{(4{)}^2}{2}$

$x=4-\frac{(4{)}^2}{4}=0m$