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

Answer:

★Position of elevator after 30 minutes will be –140 metres ★

Step-by-step explanation:

Given:

• Elevator descends at the rate of 5 m/min
• Distance above the ground is +ve
• Distance below the ground is –ve
• Elevator begins to descent from 10 metre

To Find:

• Position of elevator after 30 minutes

Solution: Initially elevator is 10 m above the ground

∴ Initial position ( +10m)

→ Now, Elevator moves down for 30 minutes

A/q ,

→Elevator moves 5 metre per minutes←

∴Distance covered by it in 30 minutes= 30 x (–5)

$\small\implies{\sf }$ –150 m below the ground

Therefore, Final position of elevator = Initial position + Distance moved in 30 minutes

$\small\implies{\sf }$ 10 + (–150)

$\small\implies{\sf }$ 10 – 150

$\small\implies{\sf }$ –140 metres

Hence, Position of elevator after 30 minutes will be –140 metres below the ground.