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 find the acceleration of object Q such that it can catch up to object P after starting 30 minutes later.

### Step 1: Determine the time taken by object P to travel 20 km

Object P starts from rest with an acceleration of \(40 \, \text{km/h}^2\). We can use the second equation of motion:

\[ S = Ut + \frac{1}{2} A t^2 \]

Where:
- \(S = 20 \, \text{km}\)
- \(U = 0 \, \text{km/h}\) (initial velocity)
- \(A = 40 \, \text{km/h}^2\)
- \(t\) is the time in hours.

Substituting the values into the equation:

\[ 20 = 0 + \frac{1}{2} \times 40 \times t^2 \]

This simplifies to:

\[ 20 = 20 t^2 \]

Dividing both sides by 20 gives:

\[ 1 = t^2 \]

Taking the square root of both sides, we find:

\[ t = 1 \, \text{hour} \]

### Step 2: Determine the time available for object Q to catch up

Since object Q starts 30 minutes (or \(0.5\) hours) after object P, it has:

\[ 1 \, \text{hour} - 0.5 \, \text{hours} = 0.5 \, \text{hours} \]

to catch up to P.

### Step 3: Use the equation of motion for object Q

Object Q also starts from rest and needs to travel the same distance of \(20 \, \text{km}\) in \(0.5 \, \text{hours}\). Using the same equation of motion:

\[ S = Ut + \frac{1}{2} A' t^2 \]

Where:
- \(S = 20 \, \text{km}\)
- \(U = 0 \, \text{km/h}\)
- \(A' = \text{acceleration of Q}\)
- \(t = 0.5 \, \text{hours}\)

Substituting the values:

\[ 20 = 0 + \frac{1}{2} A' (0.5)^2 \]

This simplifies to:

\[ 20 = \frac{1}{2} A' \times 0.25 \]

Which further simplifies to:

\[ 20 = \frac{1}{8} A' \]

Multiplying both sides by 8 gives:

\[ A' = 160 \, \text{km/h}^2 \]

### Conclusion

The acceleration of object Q required to catch up with object P is:

\[ \boxed{160 \, \text{km/h}^2} \]