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7. The distance between two places is 120 km. A car traveling at a certain speed can cover the distance in 2 hours 30 minutes. By how many km/h should the speed of the car be increased for it to take 30 minutes less to cover the distance?

8. Two cars start from the same place at the same time in the same direction. The speed of the first car is 30 m/s. How far will the second car travel if it goes 10 m/s faster than the first car over a period of 2 hours?


Sagot :

Sure, let's break down the steps to solve the problem:

Question: The distance between two places is 120 km. A car traveling at a certain speed can cover the distance in 2 hours 30 minutes. By how many km/h should the speed of the car be increased for it to take 30 minutes less to cover the distance?

1. Convert the initial time from hours and minutes to just hours:
- The initial time taken is 2 hours 30 minutes.
- 30 minutes is equal to 0.5 hours.
- Therefore, the total initial time is [tex]\(2 + 0.5 = 2.5\)[/tex] hours.

2. Calculate the initial speed:
- The distance between the two places is 120 km.
- The initial time taken is 2.5 hours.
- The speed is given by [tex]\(\text{Speed} = \frac{\text{Distance}}{\text{Time}}\)[/tex].
- So, the initial speed is [tex]\(\frac{120 \text{ km}}{2.5 \text{ hours}} = 48 \text{ km/h}\)[/tex].

3. Find the new time after reducing 30 minutes:
- The new time taken is 30 minutes less than the initial time.
- 30 minutes is equal to 0.5 hours.
- Therefore, the new time taken is [tex]\(2.5 - 0.5 = 2\)[/tex] hours.

4. Calculate the new speed:
- The distance remains the same at 120 km.
- The new time taken is 2 hours.
- The new speed is [tex]\(\frac{120 \text{ km}}{2 \text{ hours}} = 60 \text{ km/h}\)[/tex].

5. Determine the increase in speed:
- The initial speed was 48 km/h.
- The new speed is 60 km/h.
- The increase in speed is [tex]\(60 \text{ km/h} - 48 \text{ km/h} = 12 \text{ km/h}\)[/tex].

Therefore, the speed of the car should be increased by 12 km/h for it to take 30 minutes less to cover the distance of 120 km.