Respuesta :
Complete question is;
Alan leaves Los Angeles at 8:AM to drive to San Francisco, 400 mi away. He travels at a steady 47 mph. Beth leaves Los Angeles at 9:00 AM and drives a steady 55.0 mph.
a. Who gets to San Francisco first?
b. How long does the first to arrive have to wait for the second?
Answer:
A) Beth will get there first
B) Beth will have to wait for 15 minutes for alan
Explanation:
We are given;
Distance between Los Angeles and San Francisco;d = 400 mi
Alan's velocity;v_a = 47 mph
Beth's velocity;v_b = 55 mph
we know that time is given by;
Time = distance/velocity
Time required for alan;
t_a = 400/47
t_a = 8.5106 hours = 8 hours 31 minutes
Time required for Beth;
t_b = 400/55
t_b = 7.2727 hours = 7 hours 16 minutes
So Alan will reach there at;
8:00 a.m + 8 hours 31 minutes = 4:31 p.m
Beth will reach there at;
9:00 a.m + 7 hours 16 minutes = 4:16 pm
Thus, beth will arrive 4:31 pm - 4:16pm = 15 minutes before alan
A) Beth will get there first
B) Beth will have to wait for 15 minutes for alan
