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 these

Explanation

To solve the problem step by step, we will use the equations of motion under constant acceleration.

### Step 1: Identify the known values
- Distance traveled in the first 5 seconds, \( s_1 = 10 \, m \)
- Time for the first segment, \( t_1 = 5 \, s \)
- Distance traveled in the next 3 seconds, \( s_2 = 10 \, m \)
- Time for the second segment, \( t_2 = 3 \, s \)

### Step 2: Write the equations of motion
Using the equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]

For the first 5 seconds:
\[ s_1 = u t_1 + \frac{1}{2} a t_1^2 \]
Substituting the known values:
\[ 10 = u(5) + \frac{1}{2} a (5^2) \]
\[ 10 = 5u + \frac{25}{2} a \]
Multiplying through by 2 to eliminate the fraction:
\[ 20 = 10u + 25a \]
This simplifies to:
\[ 2u + 5a = 4 \quad \text{(Equation 1)} \]

### Step 3: Write the second equation for the next 3 seconds
For the next 3 seconds (from t = 5s to t = 8s):
\[ s_2 = u t_2 + \frac{1}{2} a t_2^2 \]
The total distance traveled at 8 seconds is:
\[ s_1 + s_2 = 20 = u(8) + \frac{1}{2} a (8^2) \]
Substituting the known values:
\[ 20 = 8u + \frac{1}{2} a (64) \]
\[ 20 = 8u + 32a \]
Multiplying through by 2:
\[ 40 = 16u + 64a \]
This simplifies to:
\[ 2u + 8a = 5 \quad \text{(Equation 2)} \]

### Step 4: Solve the equations simultaneously
Now we have two equations:
1. \( 2u + 5a = 4 \)
2. \( 2u + 8a = 5 \)

Subtract Equation 1 from Equation 2:
\[ (2u + 8a) - (2u + 5a) = 5 - 4 \]
This simplifies to:
\[ 3a = 1 \]
Thus, we find:
\[ a = \frac{1}{3} \, m/s^2 \]

### Step 5: Substitute \( a \) back to find \( u \)
Substituting \( a \) back into Equation 1:
\[ 2u + 5\left(\frac{1}{3}\right) = 4 \]
\[ 2u + \frac{5}{3} = 4 \]
Multiplying through by 3 to eliminate the fraction:
\[ 6u + 5 = 12 \]
Thus:
\[ 6u = 7 \]
So:
\[ u = \frac{7}{6} \, m/s \]

### Step 6: Calculate the distance traveled in the next 2 seconds (from t = 8s to t = 10s)
Using the equation of motion again for the total time of 10 seconds:
\[ s_{10} = ut + \frac{1}{2} a t^2 \]
Substituting \( t = 10 \):
\[ s_{10} = \left(\frac{7}{6}\right)(10) + \frac{1}{2}\left(\frac{1}{3}\right)(10^2) \]
Calculating:
\[ s_{10} = \frac{70}{6} + \frac{1}{2} \cdot \frac{1}{3} \cdot 100 \]
\[ s_{10} = \frac{70}{6} + \frac{50}{3} \]
Converting \( \frac{50}{3} \) to sixths:
\[ \frac{50}{3} = \frac{100}{6} \]
So:
\[ s_{10} = \frac{70}{6} + \frac{100}{6} = \frac{170}{6} \]

### Step 7: Calculate the distance traveled in the next 2 seconds
Now we need to find the distance traveled in the last 2 seconds:
\[ \Delta s = s_{10} - s_8 \]
Where \( s_8 = 20 \, m \):
\[ \Delta s = \frac{170}{6} - 20 \]
Converting 20 to sixths:
\[ 20 = \frac{120}{6} \]
Thus:
\[ \Delta s = \frac{170}{6} - \frac{120}{6} = \frac{50}{6} = \frac{25}{3} \]

### Final Answer
The distance traveled in the next 2 seconds is:
\[ \Delta s = \frac{25}{3} \approx 8.33 \, m \]

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