An elevator descends into a mine shaft at the rate of 6 m/min. If the descent starts from 10 m above the ground level, how long will it take to reach – 350 m.

Options

A. `1.5h`
B. `1hr 5min`
C. `2hr`
D. `1hr`

Explanation

To solve the problem step-by-step, we will follow the given information and apply the necessary calculations.

### Step 1: Understand the Initial and Final Positions
- The elevator starts at a height of **10 m** above the ground level.
- The final position of the elevator is **–350 m**, which means it is **350 m below** the ground level.

### Step 2: Calculate the Total Distance to be Descended
- To find the total distance the elevator needs to descend, we need to calculate the distance from the starting point (10 m) to the final point (–350 m).
- The total distance can be calculated as:
\[ \text{Total Distance} = \text{Final Position} - \text{Initial Position} = (-350) - 10 \]
\[ \text{Total Distance} = -350 - 10 = -360 \text{ m} \]

### Step 3: Determine the Rate of Descent
- The elevator descends at a rate of **6 m/min**.

### Step 4: Use the Speed Formula to Find Time
- The formula for speed is:
\[ \text{Speed} = \frac{\text{Distance}}{\text{Time}} \]
- Rearranging the formula to find time gives us:
\[ \text{Time} = \frac{\text{Distance}}{\text{Speed}} \]
- Substituting the values we have:
\[ \text{Time} = \frac{-360 \text{ m}}{-6 \text{ m/min}} \]
\[ \text{Time} = \frac{360}{6} = 60 \text{ minutes} \]

### Step 5: Conclusion
- Therefore, the time it will take for the elevator to reach –350 m is **60 minutes**, which is equivalent to **1 hour**.

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