Question

An elevator descends into a mine shaft at the rate of 6 metre per minute. What will be its position after 1 hour? If it begins to descends from 15 metre above the ground what will be its position after 40 minutes

Answer 1

${\blue{ \underline{\underline{\huge {\mathfrak \pink {Answer:-}}}}}}$

RATE=6m/min

$\mathtt{ find \: the \: distance \: travled \: below } \\ \mathtt {ground \: in \: one \: hour}.$

$: \rm\implies \: 1min = 6m \\ : \rm\implies60 min \: = 6 \times 60 = 360m$

$\mathtt \green{The \: shaft \: will \: be \: 360m \: below} \\ \mathtt \green{the \: ground \: after \: a \: hour}.$

$\mathtt{find \: the \: distance \: traveled \: in \: 40min}$

$: \rm\implies \: 1min = 4m\\ : \rm\implies40 min \: = 6 \times 40 = 240m$

$\mathtt{ find \: the \: relative \: distance \: of \: the } \\ \mathtt{shift \: after \: 40min.}$

$\bold {Given \: that \: it \: start \: from \: 15m } \\ \bold{ above \: the \: ground.}$

$\boxed{\rm{Distance = 15 - 240 \implies \: - 225}}$

$\rm \purple{It \: will \: be \: 225 \: below \: the \: ground } \\ \rm\purple{ after \: 40min}.$