Question

an elevator descends into a mine shaftat a rate of 5 metres per minutes . if we represent the distance above the ground by a positive integer and that below the ground by a negative integer and the elevator begins to descent from 10 meter above the ground then what will be the position after 30 min

Answer 1

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

$\green{\tt{\therefore{Distance=140\:m}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green {\underline \bold{Given : }} \\ \tt{: \implies Initial \: speed(u) = 5 \: m\: {min}^{ - 1} } \\ \\ \tt{: \implies Time(t) = 30 \: min} \\ \\ \red{\underline \bold{To \: Find : }} \\ \tt{: \implies Distance(s) = ?}$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt{: \implies S= ut + \frac{1}{2} a {t}^{2}} \\ \\ \tt{: \implies s + 10 = 5 \times 30 + \frac{1}{2} \times 0 \times {30}^{2}} \\ \\ \tt{: \implies s + 10 = 150} \\ \\ \tt{: \implies s = 150 - 10} \\ \\ \green{\tt{: \implies s = 140 \: m}} \\ \\ \green{\tt{\therefore The\:elevator\:descends\:140\:m\:below\:the\:ground}} \\\\ \purple{\text{Some \: formula \: related \: to \: this \: topic}} \\ \pink {\tt{ \circ \: v = u + at}} \\ \\ \pink {\tt{ \circ \: {v}^{2} = {u}^{2} + 2as}}$

Answer 2

$\huge\bold\green{Question}$

An elevator descends into a mine shaftat a rate of 5 metres per minutes . if we represent the distance above the ground by a positive integer and that below the ground by a negative integer and the elevator begins to descent from 10 meter above the ground then what will be the position after 30 min

$\huge\bold\green{Answer}$

According to the question we have given :-

Initial speed (u) = 5 metres per minutes

Time(t) = 30 min

Hence, we have to find out the distance

Distance(s) = ______ ?

So, simply by applying known formulas we can find it

$\sf\red{ S= ut + \frac{1}{2} a {t}^{2}}$

So , now by substituting the kown values in formula we get :-

$\sf{= s + 10 = 5 \times 30 + \frac{1}{2} \times 0 \times {30}^{2}} \\ \\ \sf{= s + 10 = 150} \\ \\ \sf{= s = 150 - 10} \\ \\ \sf{= s = 140 \: m} \\ \\ \sf{\Hence The\:elevator\:descends\:140\:m\:below\:the\:ground}$