1) if you live 10.0 km from your school and the bus takes 0.53 h to reach the school driving due east, what is the average velocity of the bus?
2) To qualify for the finals in a racing event, a race car must achieve an average speed of 250 km/h on a track with a total length of 1600m. if a particular car covers the first half of the track at an average speed of 230 km/h, what minimum average speed must it have in the second half of the event to qualify?
1) To find the average velocity of the bus, we need to divide the total displacement by the total time taken.
Given:
Distance to school = 10.0 km
Time taken = 0.53 h
To convert time to seconds, we need to multiply by 60 (since there are 60 minutes in an hour) and then by 60 again (since there are 60 seconds in a minute):
Total time taken = 0.53 h * 60 min/h * 60 s/min = 1908 s
The displacement of the bus is the distance to the school, which is 10.0 km, in the east direction.
Average velocity = displacement / total time taken
= 10.0 km / 1908 s
≈ 0.00524 km/s (rounded to 5 decimal places)
Therefore, the average velocity of the bus is approximately 0.00524 km/s.
2) To find the minimum average speed required in the second half of the event, we need to calculate the remaining distance and the remaining time.
Given:
Total track length = 1600 m
Average speed for first half = 230 km/h
To convert the average speed for the first half from km/h to m/s, we need to divide it by 3.6:
Average speed for the first half = 230 km/h / 3.6
≈ 63.89 m/s (rounded to 2 decimal places)
The distance covered in the first half of the track is half of the total track length, which is 1600m / 2 = 800m.
The time taken to cover the distance in the first half of the track can be calculated using the formula time = distance / speed:
Time for first half = 800 m / 63.89 m/s
≈ 12.51 s (rounded to 2 decimal places)
The remaining distance for the second half of the track is also 800m, and the remaining time is the total time minus the time taken for the first half.
Remaining distance = 800 m
Remaining time = 1908 s - 12.51 s
≈ 1895.49 s (rounded to 2 decimal places)
Now, we can calculate the minimum average speed required for the second half of the event.
Minimum average speed for the second half = Remaining distance / Remaining time
= 800 m / 1895.49 s
≈ 0.4221 m/s (rounded to 4 decimal places)
Converting the speed from m/s to km/h:
Minimum average speed for the second half ≈ 0.4221 m/s * 3.6
≈ 1.52 km/h (rounded to 2 decimal places)
Therefore, the minimum average speed required in the second half of the event to qualify is approximately 1.52 km/h.
1) To find the average velocity of the bus, we first need to know the displacement and the time taken. In this case, the displacement is the distance between your home and school, which is 10.0 km. The time taken is given as 0.53 hours.
Average velocity is calculated by dividing the displacement by the time taken:
Average velocity = displacement / time taken
Plugging in the values, we get:
Average velocity = 10.0 km / 0.53 h
To simplify this calculation, we convert the time from hours to seconds:
Average velocity = 10.0 km / (0.53 h * 3600 s/h)
Now we can convert the distance from kilometers to meters:
Average velocity = (10.0 km * 1000 m/km) / (0.53 h * 3600 s/h)
Simplifying, we find:
Average velocity = 5.917 m/s
Therefore, the average velocity of the bus is 5.917 m/s.
2) To calculate the minimum average speed required in the second half of the event, we need to find the remaining distance the car needs to cover and the time it has to do so.
The total length of the track is given as 1600m, and the first half is covered at an average speed of 230 km/h. Therefore, the distance covered in the first half of the event is 800m.
To find the remaining distance, we subtract the distance covered in the first half from the total length of the track:
Remaining distance = 1600m - 800m
= 800m
Now we need to know the time available to cover the remaining distance. For this, we need to know the average speed of the car in the first half.
Average speed is calculated by dividing the total distance covered by the time taken:
Average speed = distance covered / time taken
Since the distance covered in the first half is 800m and the average speed is given as 230 km/h, we can calculate the time taken as:
Time taken = distance covered / average speed
Time taken = 800m / (230 km/h * 1000 m/km/ 3600 s/h)
Simplifying, we find:
Time taken = 0.124 s
Therefore, the car has 0.124 seconds to cover the remaining 800m in the second half of the event.
Now we can find the minimum average speed required in the second half. The remaining distance is 800m, and the time available is 0.124 seconds.
Average speed = distance / time
Average speed = 800m / (0.124 s)
Simplifying, we find:
Average speed = 6451.6 m/s
Therefore, the car must have a minimum average speed of 6451.6 m/s in the second half of the event to qualify for the finals.
1. Velocity = Displacement / time
Displacement = Final position - initial position
Note that displacement does not have to equal distance if the bus travels in a zig-zag route.
2. Speed = distance / time
Total distance = 1600 m.
230 km/h = 230000 m/3600 s = 63.889 m/s
250 km/h = 250000 m/3600 s = 69.444 m/s
Time spent on first half of track
=800/63.889 s.
Total time permitted = 1600/69.444 s
Time left, t = 1600/69.444 - 800/63.889 s
So speed required = 800 m / t
Convert to km/h.