A person drops a stone from a building of height 20 m. At the same instant the front end of a truck passes below the building moving with constant acceleration of `1 m//s^(2)` and velocity of `2 m//s` at that instant. Length of the truck if the stone just misses to hit its rear part is :-

Options

A. `6 m`
B. `4 m`
C. `5 m`
D. `2 m`

Explanation

To solve the problem step by step, we will analyze the motion of the stone and the truck separately and then relate their motions to find the length of the truck.

### Step 1: Calculate the time of fall of the stone
The stone is dropped from a height of 20 m. We can use the formula for the time of fall under gravity:

\[ t = \sqrt{\frac{2h}{g}} \]

Where:
- \( h = 20 \, \text{m} \) (height of the building)
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)

Substituting the values:

\[ t = \sqrt{\frac{2 \times 20}{10}} = \sqrt{\frac{40}{10}} = \sqrt{4} = 2 \, \text{s} \]

### Step 2: Calculate the horizontal displacement of the truck
The truck is moving with an initial velocity \( u = 2 \, \text{m/s} \) and has a constant acceleration \( a = 1 \, \text{m/s}^2 \). We can use the equation of motion to find the displacement \( S \) of the truck during the time \( t \):

\[ S = ut + \frac{1}{2} a t^2 \]

Substituting the values:

\[ S = (2 \, \text{m/s})(2 \, \text{s}) + \frac{1}{2} (1 \, \text{m/s}^2)(2 \, \text{s})^2 \]

Calculating each term:

1. \( ut = 2 \times 2 = 4 \, \text{m} \)
2. \( \frac{1}{2} a t^2 = \frac{1}{2} \times 1 \times 4 = 2 \, \text{m} \)

Adding these together:

\[ S = 4 + 2 = 6 \, \text{m} \]

### Step 3: Determine the length of the truck
Since the stone just misses the rear part of the truck, the length of the truck must be equal to the horizontal displacement of the truck during the time the stone is falling. Therefore, the length of the truck is:

\[ \text{Length of the truck} = S = 6 \, \text{m} \]

### Final Answer
The length of the truck is **6 meters**.

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