An elevator descends into a mine shaft at the rate of 7m/min.If the descent starts from 5m above the ground level, how long will it take to reach -205m? Please answer this question. And in advance, thanks!
Answers
Answer:
Answer:
\green{\tt{\therefore{Time=70\:min}}}∴Time=70min
\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}
Step−by−stepexplanation:
\begin{gathered}\green{\underline \bold{Given : }} \\ \tt: \implies Initial \: speed = 7 \: m/min \\ \\ \tt: \implies Total \: distance = 490 \: m \\ \\ \red{\underline \bold{To \: Find : }}\\ \tt: \implies Time \: taken = ?\end{gathered}
Given:
:⟹Initialspeed=7m/min
:⟹Totaldistance=490m
ToFind:
:⟹Timetaken=?
• According to given question :
\begin{gathered} \tt \circ \: Acceleration = 0 \: {ms}^{2} \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies s = ut + \frac{1}{2} a {t}^{2} \\ \\ \tt: \implies 490 = 7 \times t + \frac{1}{2} \times 0 \times {t}^{2} \\ \\ \tt: \implies 490 = 7 \times t \\ \\ \tt: \implies t = \frac{490}{7} \\ \\ \green{\tt: \implies t = 70 \: min} \\ \\ \bold{Alternate \: method : } \\ \tt: \implies Time = \frac{Distance}{Speed} \\ \\ \tt: \implies Time = \frac{490}{7} \times \frac{metre \times min}{metre} \\ \\ \green{\tt: \implies Time = 70 \: min}\end{gathered}
∘Acceleration=0ms
2
Asweknowthat
:⟹s=ut+
2
1
at
2
:⟹490=7×t+
2
1
×0×t
2
:⟹490=7×t
:⟹t=
7
490
:⟹t=70min
Alternatemethod:
:⟹Time=
Speed
Distance
:⟹Time=
7
490
×
metre
metre×min
:⟹Time=70min