Question

1. a family on vacation in an rv leaves home at 9 a.m., travels at an average speed of 50 miles per hour, and arrives at the vacation destination at 2 p.m. with no stops along the way. at approximately what time would the family have arrived if the average speed of the trip had been 65 miles per hour?

a 12:24 p.m.
b 12:51 p.m.
c 1:24 p.m.
d 1:51 p.m.

Explanation:

Step1: Calculate original travel time

From 9 a.m. to 2 p.m., the time duration is \(2 - 9 + 12 = 5\) hours (converting 2 p.m. to 14:00 and 9 a.m. to 09:00, \(14 - 9 = 5\) hours). The original speed is 50 miles per hour, so the distance \(d = 50\times5 = 250\) miles.

Step2: Calculate new travel time

With a speed of 65 miles per hour, the new time \(t=\frac{d}{65}=\frac{250}{65}\approx3.846\) hours.

Step3: Convert decimal time to minutes and find arrival time

0.846 hours \(\times60\approx50.76\) minutes, so total time is about 3 hours and 51 minutes. Adding to 9 a.m.: 9 + 3 = 12 (noon), plus 51 minutes, so arrival time is approximately 12:51 p.m.

Answer:

C. 12:51 p.m.