A particle moves along straight line for 10 second with unifrom acceleration, `a = 2 ms^(-2)` . Then for next 10 second moves with acceleration, `a = 4 ms^(-1)`. The total displacement of particle would be (initial velocity of particle is zero)

Options

A. 400 m
B. 500 m
C. 550 m
D. 650 m

Answer

To solve the problem, we will break it down into two parts based on the two different accelerations the particle experiences during its motion.

### Step 1: Calculate the displacement during the first 10 seconds (with acceleration \( a_1 = 2 \, \text{m/s}^2 \))

Given:
- Initial velocity \( u_1 = 0 \, \text{m/s} \) (the particle starts from rest)
- Acceleration \( a_1 = 2 \, \text{m/s}^2 \)
- Time \( t_1 = 10 \, \text{s} \)

Using the second equation of motion:
\[ s_1 = u_1 t_1 + \frac{1}{2} a_1 t_1^2 \]
Substituting the values:
\[ s_1 = 0 \cdot 10 + \frac{1}{2} \cdot 2 \cdot (10)^2 \]
\[ s_1 = 0 + \frac{1}{2} \cdot 2 \cdot 100 \]
\[ s_1 = 0 + 100 = 100 \, \text{m} \]

### Step 2: Calculate the final velocity after the first 10 seconds

Using the first equation of motion:
\[ v_1 = u_1 + a_1 t_1 \]
Substituting the values:
\[ v_1 = 0 + 2 \cdot 10 \]
\[ v_1 = 20 \, \text{m/s} \]

### Step 3: Calculate the displacement during the next 10 seconds (with acceleration \( a_2 = 4 \, \text{m/s}^2 \))

Now, the initial velocity for this phase is \( u_2 = v_1 = 20 \, \text{m/s} \).
Given:
- Acceleration \( a_2 = 4 \, \text{m/s}^2 \)
- Time \( t_2 = 10 \, \text{s} \)

Using the second equation of motion again:
\[ s_2 = u_2 t_2 + \frac{1}{2} a_2 t_2^2 \]
Substituting the values:
\[ s_2 = 20 \cdot 10 + \frac{1}{2} \cdot 4 \cdot (10)^2 \]
\[ s_2 = 200 + \frac{1}{2} \cdot 4 \cdot 100 \]
\[ s_2 = 200 + 200 = 400 \, \text{m} \]

### Step 4: Calculate the total displacement

Now, we can find the total displacement \( S \) by adding the displacements from both phases:
\[ S = s_1 + s_2 \]
\[ S = 100 + 400 = 500 \, \text{m} \]

Thus, the total displacement of the particle is \( \boxed{500 \, \text{m}} \).