A particle travels 10m in first 5 sec and 10 m in next 3 sec. Assuming constant acceleration what is the distance travelled in next 2 sec.

Options

A. 8.3 m
B. 9.3 m
C. 10.3 m
D. none of above

Explanation

To solve the problem step by step, we will use the equations of motion under constant acceleration. Here’s how we can approach the problem:

### Step 1: Understand the problem
We know that a particle travels 10 meters in the first 5 seconds and another 10 meters in the next 3 seconds. We need to find the distance it travels in the next 2 seconds, assuming constant acceleration.

### Step 2: Define the variables
Let:
- \( u \) = initial velocity (we will assume it is 0 for simplicity)
- \( a \) = constant acceleration
- \( s_1 \) = distance traveled in the first 5 seconds = 10 m
- \( s_2 \) = distance traveled in the next 3 seconds = 10 m
- \( s_3 \) = distance traveled in the next 2 seconds (what we need to find)

### Step 3: Use the equations of motion
The equation of motion we will use is:
\[ s = ut + \frac{1}{2} a t^2 \]

#### For the first 5 seconds:
Using the equation for the first 5 seconds:
\[ s_1 = u \cdot 5 + \frac{1}{2} a (5^2) \]
Substituting \( s_1 = 10 \):
\[ 10 = 5u + \frac{25}{2} a \quad \text{(Equation 1)} \]

#### For the first 8 seconds:
In the next 3 seconds, the total time is now 8 seconds:
\[ s_2 = u \cdot 8 + \frac{1}{2} a (8^2) \]
The total distance traveled in 8 seconds is \( s_1 + s_2 = 20 \):
\[ 20 = 8u + \frac{1}{2} a (64) \]
This simplifies to:
\[ 20 = 8u + 32a \quad \text{(Equation 2)} \]

### Step 4: Solve the equations
Now we have two equations:
1. \( 10 = 5u + \frac{25}{2} a \)
2. \( 20 = 8u + 32a \)

We can solve these equations simultaneously.

#### Rearranging Equation 1:
From Equation 1:
\[ 5u + \frac{25}{2} a = 10 \rightarrow 2u + 5a = 4 \quad \text{(Equation 3)} \]

#### Rearranging Equation 2:
From Equation 2:
\[ 8u + 32a = 20 \rightarrow 2u + 8a = 5 \quad \text{(Equation 4)} \]

### Step 5: Subtract Equation 3 from Equation 4
Subtracting Equation 3 from Equation 4:
\[ (2u + 8a) - (2u + 5a) = 5 - 4 \]
This simplifies to:
\[ 3a = 1 \rightarrow a = \frac{1}{3} \, \text{m/s}^2 \]

### Step 6: Substitute \( a \) back to find \( u \)
Now substituting \( a \) back into Equation 3:
\[ 2u + 5\left(\frac{1}{3}\right) = 4 \]
\[ 2u + \frac{5}{3} = 4 \]
\[ 2u = 4 - \frac{5}{3} = \frac{12}{3} - \frac{5}{3} = \frac{7}{3} \]
\[ u = \frac{7}{6} \, \text{m/s} \]

### Step 7: Calculate the distance traveled in the next 2 seconds
Now we calculate the distance traveled in the next 2 seconds (from 8 seconds to 10 seconds):
Using the equation:
\[ s_3 = u \cdot t + \frac{1}{2} a t^2 \]
Where \( t = 2 \) seconds:
\[ s_3 = \left(\frac{7}{6}\right)(2) + \frac{1}{2}\left(\frac{1}{3}\right)(2^2) \]
\[ s_3 = \frac{14}{6} + \frac{1}{2}\left(\frac{1}{3}\right)(4) \]
\[ s_3 = \frac{14}{6} + \frac{2}{3} \]
Converting \( \frac{14}{6} \) to a common denominator of 3:
\[ s_3 = \frac{14}{6} + \frac{4}{6} = \frac{18}{6} = 3 \, \text{m} \]

### Step 8: Total distance traveled in 10 seconds
Now, the total distance traveled in 10 seconds is:
\[ s_{10} = s_1 + s_2 + s_3 = 10 + 10 + 3 = 23 \, \text{m} \]

### Step 9: Calculate the distance in the last 2 seconds
The distance traveled in the last 2 seconds is:
\[ s_{last} = s_{10} - s_2 = 23 - 20 = 3 \, \text{m} \]

### Final Answer
The distance traveled in the next 2 seconds is **3 meters**.