The Cedar Bluff express KAT bus averages 15 mph for 20 minutes, and then 32 mph for 14 minutes. What was its average speed? (mph)
d = V1*t1 + V2*t21,
d = 15 mi/h*(20/60)h + 32mi/h*(14/60)h,
d = 5 + 7.5 = 12.5 Miles.
t=20 min. + 14 min = 34 min = 0.567 h.
V(ave) = d / t = 12.5 / 0.567 = 22.1 mi/h.
thanks, I see where I messed up at...
To find the average speed, you need to calculate the total distance traveled and divide it by the total time taken.
First, let's convert the time values into hours, as average speed is typically measured in miles per hour (mph).
The bus travels at an average speed of 15 mph for 20 minutes. To convert minutes into hours, divide 20 by 60 (since there are 60 minutes in an hour):
20 minutes / 60 = 0.333 hours
Similarly, the bus travels at an average speed of 32 mph for 14 minutes. Converting minutes to hours:
14 minutes / 60 = 0.233 hours
Now let's calculate the total distance traveled.
Distance = Speed × Time
Distance for the first leg of the journey:
Distance1 = 15 mph × 0.333 hours
Distance1 = 4.995 miles (rounded to three decimal places)
Distance for the second leg of the journey:
Distance2 = 32 mph × 0.233 hours
Distance2 = 7.456 miles (rounded to three decimal places)
To calculate the total distance, sum Distance1 and Distance2:
Total Distance = Distance1 + Distance2
Total Distance = 4.995 miles + 7.456 miles
Total Distance = 12.451 miles (rounded to three decimal places)
Now we can calculate the average speed:
Average Speed = Total Distance / Total Time
Total Time = (20 minutes + 14 minutes) / 60 (to convert to hours)
Total Time = 34 minutes / 60 = 0.567 hours
Average Speed = 12.451 miles / 0.567 hours
Average Speed = 21.93 mph (rounded to two decimal places)
Therefore, the average speed of the Cedar Bluff express KAT bus is approximately 21.93 mph.