A particle moves in a straight line with a constant acceleration. It changes its velocity from `10 ms^-1` to `20 ms^-1` while passing through a distance `135 m` in `t` seconds. The value of `t` is.

Options

A. 10
B. 1.8
C. 12
D. 9

Explanation

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

### Step 1: Identify the known values
- Initial velocity (u) = 10 m/s
- Final velocity (v) = 20 m/s
- Distance (s) = 135 m

### Step 2: Use the equation of motion to find acceleration (a)
We can use the equation:
\[ v^2 = u^2 + 2as \]
Substituting the known values:
\[ (20)^2 = (10)^2 + 2a(135) \]

### Step 3: Simplify the equation
Calculating the squares:
\[ 400 = 100 + 270a \]
Now, subtract 100 from both sides:
\[ 400 - 100 = 270a \]
\[ 300 = 270a \]

### Step 4: Solve for acceleration (a)
Now, divide both sides by 270:
\[ a = \frac{300}{270} \]
\[ a = \frac{10}{9} \, \text{m/s}^2 \]

### Step 5: Use the acceleration to find time (t)
Now we can use the equation:
\[ v = u + at \]
Substituting the known values:
\[ 20 = 10 + \left(\frac{10}{9}\right)t \]

### Step 6: Rearrange to solve for time (t)
Subtract 10 from both sides:
\[ 20 - 10 = \left(\frac{10}{9}\right)t \]
\[ 10 = \left(\frac{10}{9}\right)t \]

### Step 7: Solve for t
Now, multiply both sides by \( \frac{9}{10} \):
\[ t = 10 \times \frac{9}{10} \]
\[ t = 9 \, \text{seconds} \]

### Conclusion
The time taken (t) for the particle to change its velocity from 10 m/s to 20 m/s while covering a distance of 135 m is **9 seconds**.

---