You are designing an elevator for a hospital. The force exerted on a passenger by the floor of the elevator is not to exceed 1.60 times the passenger's weight. The elevator accelerates upwards with constant accelerates upwards with constant acceleration for a distance of 3.0 m and then starts to slow down. What is the maximum speed (in `ms^(-1)`) of the elevator?

Explanation

To solve the problem step by step, we will follow these steps:

### Step 1: Understand the forces acting on the passenger
The forces acting on the passenger in the elevator are:
- The weight of the passenger acting downwards, which is \( W = mg \) (where \( m \) is the mass of the passenger and \( g \) is the acceleration due to gravity).
- The normal force \( N \) exerted by the floor of the elevator acting upwards.

### Step 2: Set up the inequality for the normal force
According to the problem, the normal force \( N \) should not exceed \( 1.6 \) times the weight of the passenger:
\[ N \leq 1.6 mg \]

### Step 3: Apply Newton's second law
When the elevator accelerates upwards, we can apply Newton's second law:
\[ N - mg = ma \]
Where \( a \) is the acceleration of the elevator. Rearranging gives:
\[ N = mg + ma \]
Substituting the inequality for \( N \):
\[ mg + ma \leq 1.6 mg \]

### Step 4: Solve for the maximum acceleration
Rearranging the inequality:
\[ ma \leq 1.6 mg - mg \]
\[ ma \leq 0.6 mg \]
Dividing both sides by \( m \) (assuming \( m \neq 0 \)):
\[ a \leq 0.6g \]

### Step 5: Substitute the value of \( g \)
Using \( g \approx 9.8 \, \text{m/s}^2 \):
\[ a \leq 0.6 \times 9.8 \, \text{m/s}^2 \]
\[ a \leq 5.88 \, \text{m/s}^2 \]

### Step 6: Calculate the maximum speed after accelerating for 3 meters
Using the kinematic equation:
\[ v^2 = u^2 + 2as \]
Where:
- \( v \) is the final velocity,
- \( u \) is the initial velocity (which is \( 0 \) since the elevator starts from rest),
- \( a \) is the maximum acceleration (\( 5.88 \, \text{m/s}^2 \)),
- \( s \) is the distance (\( 3 \, \text{m} \)).

Substituting the values:
\[ v^2 = 0 + 2 \times 5.88 \times 3 \]
\[ v^2 = 35.28 \]
\[ v = \sqrt{35.28} \]
\[ v \approx 5.94 \, \text{m/s} \]

### Final Answer:
The maximum speed of the elevator is approximately \( 5.94 \, \text{m/s} \).

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