Let the distance be \( D = 60 \) km.

Let A's usual speed be \( S_A \) km/hr and B's speed be \( S_B \) km/hr.

The time taken is calculated using the formula: \( \text{Time} = \frac{\text{Distance}}{\text{Speed}} \).

Setting Up Speed and Time Equations

Based on the problem statement:

• Initially, A takes 6 hours more than B:
  \( T_A = T_B + 6 \)
  \( \frac{60}{S_A} = \frac{60}{S_B} + 6 \) (Equation 1)
• When A's speed is doubled (\( 2S_A \)), A takes 3 hours less than B:
  \( T_{A_{new}} = T_B - 3 \)
  \( \frac{60}{2S_A} = \frac{60}{S_B} - 3 \)
  \( \frac{30}{S_A} = \frac{60}{S_B} - 3 \) (Equation 2)

Calculating A's Speed

We can solve these two equations to find \( S_A \).

1. From Equation 1, isolate the term related to B's time:
   \( \frac{60}{S_B} = \frac{60}{S_A} - 6 \)
2. Substitute this expression for \( \frac{60}{S_B} \) into Equation 2:
   \( \frac{30}{S_A} = \left( \frac{60}{S_A} - 6 \right) - 3 \)
3. Simplify the equation:
   \( \frac{30}{S_A} = \frac{60}{S_A} - 9 \)
4. Rearrange the terms to solve for \( S_A \):
   \( 9 = \frac{60}{S_A} - \frac{30}{S_A} \)
   \( 9 = \frac{30}{S_A} \)
5. Calculate the value of \( S_A \):
   \( S_A = \frac{30}{9} \)
   \( S_A = \frac{10}{3} \) km/hr
6. Express the speed as a mixed number:
   \( S_A = 3\frac{1}{3} \) km/hr

Conclusion

A's speed is \(3\frac{1}{3}\) km/hr.