Two cars P and Q are 60 km apart. They are travelling in the same direction. The constant speed of car P is 8 km/h slower than twice that of car Q. If car P catches up with car Q a... Two cars P and Q are 60 km apart. They are travelling in the same direction. The constant speed of car P is 8 km/h slower than twice that of car Q. If car P catches up with car Q after 36 minutes, find the constant speed of car Q. Read more

Steps to Solve

1. Define the variables Let the speed of car Q be $x$ km/h. Then, the speed of car P is $2x - 8$ km/h because it is 8 km/h slower than twice the speed of Q.

2. Convert time to hours The time taken for car P to catch up with car Q is 36 minutes. To convert this to hours, we have: $$ \text{Time in hours} = \frac{36}{60} = 0.6 \text{ hours} $$

3. Set up the distance equation In the 0.6 hours, car P travels the distance of 60 km plus the distance that car Q covers. The equations for distances traveled by both cars are:

• Distance traveled by P: $$ \text{Distance}_P = (2x - 8) \times 0.6 $$
• Distance traveled by Q: $$ \text{Distance}_Q = x \times 0.6 $$

4. Equate the distances Since car P catches up with car Q, the distance covered by car P equals the distance covered by car Q plus the initial distance of 60 km: $$ (2x - 8) \times 0.6 = x \times 0.6 + 60 $$

5. Distribute and simplify Distributing both sides, we get: $$ 1.2x - 4.8 = 0.6x + 60 $$ Now, combine like terms: $$ 1.2x - 0.6x = 60 + 4.8 $$ This simplifies to: $$ 0.6x = 64.8 $$

6. Solve for x Now, divide both sides by 0.6: $$ x = \frac{64.8}{0.6} = 108 \text{ km/h} $$