Two objects P and Q, travelling in the same direction start from rest. While the object P starts at time t = 0 and object Q starts later at t = 30 min. The object P has an acceleration of `40 km//h^(2)`. To catch P at a distance of 20 km, the acceleration of Q should be

Options

A. `40 km//h^(2) `
B. `80 km//h^(2)`
C. `100 km//h^(2)`
D. `160 km//h^(2)`

Answer

To solve the problem, we need to determine the acceleration of object Q so that it can catch up to object P after starting 30 minutes later.

### Step-by-Step Solution:

1. **Understanding the Problem:**
- Object P starts from rest at time \( t = 0 \) with an acceleration of \( 40 \, \text{km/h}^2 \).
- Object Q starts from rest at \( t = 30 \) minutes (or \( 0.5 \) hours) and needs to catch up to P at a distance of \( 20 \, \text{km} \).

2. **Finding the Time Taken by Object P to Cover 20 km:**
- We can use the equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
- Since object P starts from rest, \( u = 0 \), so the equation simplifies to:
\[ s = \frac{1}{2} a t^2 \]
- Rearranging for \( t \):
\[ t = \sqrt{\frac{2s}{a}} \]
- Substituting \( s = 20 \, \text{km} \) and \( a = 40 \, \text{km/h}^2 \):
\[ t = \sqrt{\frac{2 \times 20}{40}} = \sqrt{1} = 1 \, \text{hour} \]

3. **Calculating the Time Available for Object Q:**
- Since object Q starts 30 minutes (or \( 0.5 \) hours) later, it has:
\[ 1 \, \text{hour} - 0.5 \, \text{hours} = 0.5 \, \text{hours} \]
- Therefore, object Q has \( 0.5 \) hours to cover the same distance of \( 20 \, \text{km} \).

4. **Finding the Required Acceleration of Object Q:**
- Using the same equation of motion for object Q:
\[ s = ut + \frac{1}{2} a t^2 \]
- Again, since Q starts from rest, \( u = 0 \):
\[ s = \frac{1}{2} a t^2 \]
- Rearranging for \( a \):
\[ a = \frac{2s}{t^2} \]
- Substituting \( s = 20 \, \text{km} \) and \( t = 0.5 \, \text{hours} \):
\[ a = \frac{2 \times 20}{(0.5)^2} = \frac{40}{0.25} = 160 \, \text{km/h}^2 \]

5. **Conclusion:**
- The required acceleration of object Q to catch up with object P at a distance of \( 20 \, \text{km} \) is \( 160 \, \text{km/h}^2 \).

### Final Answer:
The acceleration of Q should be \( 160 \, \text{km/h}^2 \).