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.

### Step 1: Understand the problem
We know that the particle travels 10 meters in the first 5 seconds and another 10 meters in the next 3 seconds. We need to find out how far it travels in the next 2 seconds.

### Step 2: Set up the equations
We will use the second equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
where:
- \(s\) is the displacement,
- \(u\) is the initial velocity,
- \(a\) is the acceleration,
- \(t\) is the time.

### Step 3: Calculate the parameters for the first 5 seconds
For the first 5 seconds:
- Displacement \(s_1 = 10 \, \text{m}\)
- Time \(t_1 = 5 \, \text{s}\)

Using the equation:
\[ 10 = u \cdot 5 + \frac{1}{2} a \cdot (5^2) \]
This simplifies to:
\[ 10 = 5u + \frac{25a}{2} \]
Multiplying through by 2 to eliminate the fraction:
\[ 20 = 10u + 25a \quad \text{(Equation 1)} \]

### Step 4: Calculate the parameters for the next 3 seconds
For the next 3 seconds (from 5 to 8 seconds):
- Displacement \(s_2 = 10 \, \text{m}\)
- Time \(t_2 = 3 \, \text{s}\)

The total time at this point is \(5 + 3 = 8 \, \text{s}\). The displacement from \(t = 0\) to \(t = 8\) is:
\[ s = u \cdot 8 + \frac{1}{2} a \cdot (8^2) \]
This gives:
\[ 20 = 8u + \frac{1}{2} a \cdot 64 \]
Simplifying:
\[ 20 = 8u + 32a \quad \text{(Equation 2)} \]

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

We can solve these equations simultaneously. From Equation 1:
\[ u = \frac{20 - 25a}{10} \]

Substituting \(u\) in Equation 2:
\[ 8\left(\frac{20 - 25a}{10}\right) + 32a = 20 \]
Multiplying through by 10 to eliminate the fraction:
\[ 8(20 - 25a) + 320a = 200 \]
Expanding:
\[ 160 - 200a + 320a = 200 \]
Combining like terms:
\[ 120a = 40 \implies a = \frac{1}{3} \, \text{m/s}^2 \]

Now substituting \(a\) back into Equation 1 to find \(u\):
\[ 10u + 25\left(\frac{1}{3}\right) = 20 \]
\[ 10u + \frac{25}{3} = 20 \]
Multiplying through by 3:
\[ 30u + 25 = 60 \implies 30u = 35 \implies u = \frac{7}{6} \, \text{m/s} \]

### Step 6: Calculate the distance traveled in the next 2 seconds
Now we need to find the distance traveled in the next 2 seconds (from \(t = 8\) to \(t = 10\)):
Using the equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
For \(t = 10\):
\[ s_{10} = u \cdot 10 + \frac{1}{2} a \cdot (10^2) \]
Substituting \(u\) and \(a\):
\[ s_{10} = \left(\frac{7}{6}\right) \cdot 10 + \frac{1}{2} \cdot \left(\frac{1}{3}\right) \cdot 100 \]
Calculating:
\[ s_{10} = \frac{70}{6} + \frac{50}{3} \]
Converting \(\frac{50}{3}\) to sixths:
\[ s_{10} = \frac{70}{6} + \frac{100}{6} = \frac{170}{6} = \frac{85}{3} \approx 28.33 \, \text{m} \]

### Step 7: Calculate the distance traveled in the next 2 seconds
The distance traveled in the first 8 seconds is \(20 \, \text{m}\). Thus, the distance traveled in the next 2 seconds is:
\[ s_{next} = s_{10} - s_{8} = \frac{85}{3} - 20 = \frac{85}{3} - \frac{60}{3} = \frac{25}{3} \approx 8.33 \, \text{m} \]

### Final Answer
The distance traveled in the next 2 seconds is approximately \(8.33 \, \text{m}\).