A car is to be hoisted by elevator to the fourth floor of a parking garage, which is 14 m above the ground. If the elevator can have maximum acceleration of `0.2 m//s^2` and maximum deceleration of `0.1 m//s^2` and can reach a maximum speed of 2.5 m//s, determine the shortest time to make the lift, starting from rest and ending at rest.

Explanation

To solve the problem of determining the shortest time for the elevator to hoist the car to the fourth floor, we will follow these steps:

### Step 1: Understand the problem and given data
- The height to be covered (h) = 14 m
- Maximum acceleration (a) = 0.2 m/s²
- Maximum deceleration (d) = 0.1 m/s²
- Maximum speed (v_max) = 2.5 m/s

### Step 2: Determine the time to reach maximum speed
Using the equation of motion:
\[ v = u + at \]
Where:
- \( v \) = final velocity (2.5 m/s)
- \( u \) = initial velocity (0 m/s)
- \( a \) = acceleration (0.2 m/s²)

Rearranging the equation to find time (t):
\[ 2.5 = 0 + 0.2t \]
\[ t = \frac{2.5}{0.2} = 12.5 \text{ seconds} \]

### Step 3: Calculate the distance covered during acceleration
Using the formula for distance covered under uniform acceleration:
\[ s = ut + \frac{1}{2}at^2 \]
Substituting the values:
\[ s = 0 + \frac{1}{2} \cdot 0.2 \cdot t^2 \]
\[ s = 0.1t^2 \]

### Step 4: Calculate the time to decelerate from maximum speed to rest
Using the same formula:
\[ v = u + at \]
Where:
- \( v \) = final velocity (0 m/s)
- \( u \) = initial velocity (2.5 m/s)
- \( d \) = deceleration (-0.1 m/s²)

Rearranging to find time (t_d):
\[ 0 = 2.5 - 0.1t_d \]
\[ t_d = \frac{2.5}{0.1} = 25 \text{ seconds} \]

### Step 5: Calculate the distance covered during deceleration
Using the same distance formula:
\[ s_d = ut + \frac{1}{2}at^2 \]
Substituting the values:
\[ s_d = 2.5t_d - \frac{1}{2} \cdot 0.1 \cdot t_d^2 \]
\[ s_d = 2.5t_d - 0.05t_d^2 \]

### Step 6: Total distance covered
The total distance covered during acceleration and deceleration must equal the height of the parking garage:
\[ h = s_a + s_d \]
\[ 14 = 0.1t^2 + (2.5t_d - 0.05t_d^2) \]

### Step 7: Solve for total time
Since the elevator starts and ends at rest, the time taken to accelerate and decelerate will be equal. Let \( t_a = t_d = t \):
\[ 14 = 0.1t^2 + (2.5t - 0.05t^2) \]
Combining terms:
\[ 14 = 2.5t + 0.1t^2 - 0.05t^2 \]
\[ 14 = 2.5t + 0.05t^2 \]

### Step 8: Rearranging the equation
Rearranging gives us:
\[ 0.05t^2 + 2.5t - 14 = 0 \]

### Step 9: Solve the quadratic equation
Using the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Where:
- \( a = 0.05 \)
- \( b = 2.5 \)
- \( c = -14 \)

Calculating the discriminant:
\[ b^2 - 4ac = (2.5)^2 - 4 \cdot 0.05 \cdot (-14) \]
\[ = 6.25 + 2.8 = 9.05 \]

Calculating \( t \):
\[ t = \frac{-2.5 \pm \sqrt{9.05}}{2 \cdot 0.05} \]
Calculating the positive root:
\[ t = \frac{-2.5 + 3.01}{0.1} \approx 5.1 \text{ seconds} \]

### Step 10: Verify maximum speed condition
Calculating maximum speed:
\[ v = 0.2t = 0.2 \times 5.1 = 1.02 \text{ m/s} \]
Since \( 1.02 < 2.5 \), the condition is satisfied.

### Final Answer
The shortest time to make the lift is approximately **5.1 seconds**.

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