A particle is projected vertically upwards with a velocity of 20m/sec. Find the time at which distance travelled is twice the displacement 

Options

A. `2 + sqrt(4//3) sec`
B. `1 sec`
C. `2 + sqrt(3//4) sec`
D. `3 sec `

Answer

To solve the problem, we need to find the time at which the distance traveled by the particle is twice the displacement from the starting point. Let's break down the solution step by step.

### Step 1: Understand the motion of the particle
When a particle is projected vertically upwards, it will rise to a certain height before coming to a stop and then falling back down. The total distance traveled is the sum of the distance going up and the distance coming down.

### Step 2: Define the variables
- Initial velocity (u) = 20 m/s
- Acceleration due to gravity (g) = 10 m/s² (acting downwards)
- Let the time taken to reach the maximum height be \( t_1 \).
- Let the time taken to return to the original position from the maximum height be \( t_2 \).

### Step 3: Calculate the time to reach maximum height
At maximum height, the final velocity (v) will be 0. Using the equation of motion:
\[ v = u - gt_1 \]
Setting \( v = 0 \):
\[ 0 = 20 - 10t_1 \]
Solving for \( t_1 \):
\[ 10t_1 = 20 \]
\[ t_1 = 2 \text{ seconds} \]

### Step 4: Calculate the maximum height reached
Using the equation:
\[ h = ut_1 - \frac{1}{2}gt_1^2 \]
Substituting the values:
\[ h = 20 \times 2 - \frac{1}{2} \times 10 \times (2)^2 \]
\[ h = 40 - 20 = 20 \text{ meters} \]

### Step 5: Calculate the total time of flight
The total time of flight \( T \) is the time to go up and the time to come down:
\[ T = t_1 + t_2 \]
Since the time to go up is equal to the time to come down:
\[ t_2 = t_1 = 2 \text{ seconds} \]
Thus,
\[ T = 2 + 2 = 4 \text{ seconds} \]

### Step 6: Find the distance traveled and displacement
- The total distance traveled when the particle returns to the starting point is:
\[ \text{Distance traveled} = h + h = 20 + 20 = 40 \text{ meters} \]
- The displacement from the starting point when it returns is:
\[ \text{Displacement} = 0 \text{ meters} \]

### Step 7: Set up the equation for distance traveled being twice the displacement
Let \( t \) be the time at which the distance traveled is twice the displacement:
\[ \text{Distance traveled} = 2 \times \text{Displacement} \]
From the equations of motion, we can express the distance traveled and displacement at any time \( t \).

### Step 8: Solve for time \( t \)
Using the equations:
- Distance traveled up to time \( t \):
\[ d = ut - \frac{1}{2}gt^2 \]
- Displacement at time \( t \):
\[ s = ut - \frac{1}{2}gt^2 \]

Setting the condition:
\[ d = 2s \]
This gives us:
\[ ut - \frac{1}{2}gt^2 = 2(ut - \frac{1}{2}gt^2) \]
This leads to a quadratic equation that can be solved for \( t \).

### Final Solution
After solving the quadratic equation, we find the time \( t \) at which the distance traveled is twice the displacement.