Question

1. An elevator descends into a mine shaft at the rate of 6 m/min. The descent starts
10 m above sea level.
(1) Where will it reach in 20 minutes?
(11) How long will it take to reach a depth of -350 m?​

Answer 1

$\bold\red{\underline{\underline{Answer:}}}$

$\bold\green{\underline{\underline{Solution}}}$

Here,

Initial velocity(u)=0

Final velocity(v)=6m/min

v=u+at

6=0+a(1)

a=6m/$\bold{s^{2}}$

(i)$\bold{s=ut+ \frac{1}{2}×at^{2}}$

$\bold{s=3×20^{2}}$

$\bold{s=3×400}$

$\bold{s=1200m}$

It will cover distance of 1200 metres in 20 minutes.But it is 10m above sea level.

Therefore it will be at the of 1990m

(ii)Total distance to cover=350+10=360m

$\bold{s=ut+\frac{1}{2}×at^{2}}$

$\bold{360=3×t^{2}}$

$\bold{t^{2}=\frac{360}{3}}$

$\bold{t^{2}=120}$

$\bold{t=\sqrt120}$

$\bold{t=\sqrt(2×3×2×5×2)}$

$\bold{t=2\sqrt30 \ minutes}$

It will take $\bold{2\sqrt30 \ minutes}$ to reach a depth of -350 m