A particle starts from rest, accelerates at `2(m)/(s^2)` for 10 s and then goes for constant speed for 30 s and then decelerates at `4(m)/(s^2)`. Till it stops. What is the distance travelled by it?

Options

A. 750 m
B. 800 m
C. 700 m
D. 850 m

Explanation

To solve the problem of the particle's motion, we will break it down into three distinct phases: acceleration, constant speed, and deceleration. We will calculate the distance traveled in each phase and then sum them up to find the total distance.

### Step 1: Calculate the distance during acceleration (S1)
The particle starts from rest (initial velocity, \( U = 0 \)) and accelerates at \( a = 2 \, \text{m/s}^2 \) for \( t = 10 \, \text{s} \).

Using the equation of motion:
\[ S_1 = U t + \frac{1}{2} a t^2 \]
Substituting the values:
\[ S_1 = 0 \cdot 10 + \frac{1}{2} \cdot 2 \cdot (10)^2 \]
\[ S_1 = 0 + \frac{1}{2} \cdot 2 \cdot 100 = 100 \, \text{m} \]

### Step 2: Calculate the distance during constant speed (S2)
After 10 seconds, we need to find the final velocity (\( V \)) at the end of the acceleration phase:
\[ V = U + a t = 0 + 2 \cdot 10 = 20 \, \text{m/s} \]

The particle then moves at this constant speed for \( t = 30 \, \text{s} \). The distance traveled during this phase is:
\[ S_2 = V \cdot t = 20 \cdot 30 = 600 \, \text{m} \]

### Step 3: Calculate the distance during deceleration (S3)
The particle decelerates at \( a = -4 \, \text{m/s}^2 \) until it stops. The initial velocity for this phase is \( V = 20 \, \text{m/s} \) and the final velocity is \( 0 \, \text{m/s} \).

Using the equation of motion:
\[ V^2 = U^2 + 2aS \]
Rearranging gives:
\[ 0 = (20)^2 + 2(-4)S_3 \]
\[ 0 = 400 - 8S_3 \]
\[ 8S_3 = 400 \implies S_3 = \frac{400}{8} = 50 \, \text{m} \]

### Step 4: Calculate the total distance traveled
Now, we sum the distances from all three phases:
\[ \text{Total Distance} = S_1 + S_2 + S_3 = 100 + 600 + 50 = 750 \, \text{m} \]

### Final Answer
The total distance traveled by the particle is \( \boxed{750 \, \text{m}} \).

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