A car starts with constant acceleration `a=2m//s^(2)` at t=0. two coins are released from the car at t=3 and t=4 . Each coin takes 1 second to fall on ground. Then the distance between the two coins will be (Assume coin sticks to the ground)

Options

A. 9m
B. 7m
C. 15 m
D. 2m

Explanation

To solve the problem, we need to analyze the motion of the car and the coins dropped from it. Let's break it down step by step.

### Step 1: Calculate the velocity of the car at t = 3 seconds
The formula for velocity under constant acceleration is:
\[ V = U + aT \]
Where:
- \( U = 0 \) (initial velocity, since the car starts from rest)
- \( a = 2 \, \text{m/s}^2 \) (constant acceleration)
- \( T = 3 \, \text{s} \)

Substituting the values:
\[ V_1 = 0 + 2 \times 3 = 6 \, \text{m/s} \]

### Step 2: Calculate the velocity of the car at t = 4 seconds
Using the same formula:
\[ V_2 = U + aT \]
Where:
- \( T = 4 \, \text{s} \)

Substituting the values:
\[ V_2 = 0 + 2 \times 4 = 8 \, \text{m/s} \]

### Step 3: Determine the horizontal distance traveled by each coin after being dropped
When the first coin is dropped at \( t = 3 \, \text{s} \), it has a horizontal velocity of \( 6 \, \text{m/s} \). It will fall for \( 1 \, \text{s} \) before hitting the ground:
\[ \text{Distance traveled by Coin 1} (S_1) = \text{velocity} \times \text{time} = 6 \, \text{m/s} \times 1 \, \text{s} = 6 \, \text{m} \]

When the second coin is dropped at \( t = 4 \, \text{s} \), it has a horizontal velocity of \( 8 \, \text{m/s} \) and also falls for \( 1 \, \text{s} \):
\[ \text{Distance traveled by Coin 2} (S_2) = 8 \, \text{m/s} \times 1 \, \text{s} = 8 \, \text{m} \]

### Step 4: Calculate the distance traveled by the car between the two drops
The car is accelerating, so we can find the distance it travels between \( t = 3 \, \text{s} \) and \( t = 4 \, \text{s} \) using the formula:
\[ S = V_1 \cdot t + \frac{1}{2} a t^2 \]
Where:
- \( V_1 = 6 \, \text{m/s} \) (initial velocity at \( t = 3 \, \text{s} \))
- \( t = 1 \, \text{s} \) (time interval)
- \( a = 2 \, \text{m/s}^2 \)

Calculating the distance:
\[ S = 6 \cdot 1 + \frac{1}{2} \cdot 2 \cdot (1)^2 = 6 + 1 = 7 \, \text{m} \]

### Step 5: Find the distance between the two coins
Now we can find the distance between the two coins. The first coin travels \( 6 \, \text{m} \) from its drop point, and the second coin travels \( 8 \, \text{m} \) from its drop point. The car itself moves \( 7 \, \text{m} \) during the time between the drops.

The distance between the two coins is:
\[ \text{Distance} = S_2 - S_1 + \text{Distance traveled by car} \]
\[ \text{Distance} = 8 \, \text{m} - 6 \, \text{m} + 7 \, \text{m} = 9 \, \text{m} \]

### Final Answer
The distance between the two coins when they hit the ground is **9 meters**.