Sue left her home at 6 A.M. driving her car at the rate of 50 miles per hour. At 8 A.M. her brother Kevin drove after her along the same highway, traveling at the rate of 60 miles per hour, in how many hours did Kevin pass Sue.
It took Kevin 10 hours to pass Sue, given there was no stopping involved.
Sue had a 100 mile lead over Kevin, but Kevin's car goes 10 miles faster. 100/10 = 10, so 10 hours Ans.
50(2) + 50x = 60x
100 = 10x
10 = x
10 hrs
To find out in how many hours Kevin passed Sue, we need to determine the time difference between when Sue left her home and when Kevin caught up with her.
Let's calculate the time it took Sue to travel before Kevin started driving:
Distance = Rate × Time
Sue's speed = 50 miles per hour
Time = Distance ÷ Rate
Since Sue started at 6 A.M. and Kevin started at 8 A.M., the time difference is 2 hours. So, Sue had a head start of 2 hours.
Now, let's determine how far Sue traveled during this 2-hour head start:
Distance = Rate × Time
Distance = 50 miles per hour × 2 hours = 100 miles.
By the time Kevin started driving, Sue was 100 miles ahead.
To calculate the time it took Kevin to catch up with Sue:
Relative speed = Kevin's speed - Sue's speed
Relative speed = 60 miles per hour - 50 miles per hour = 10 miles per hour
Now, we can find the time it took Kevin to catch up with Sue:
Time = Distance ÷ Relative speed
Time = 100 miles ÷ 10 miles per hour = 10 hours
Therefore, Kevin caught up with Sue in 10 hours.