Dr. Worm leaves his house at exactly 7:20 a.m. every morning. When he averages 45 miles per hour, he arrives at his workplace five minutes late. When he averages 63 miles per hour, he arrives five minutes early. What speed should Dr. Worm average to arrive at his workplace precisely on time?
+285 Let the distance from Dr. Lease's home to his workplace be d and the desired travel time be t. We have the two equations:
d/45=t+5
d/63=t-5
Adding and solving for distance divided by time (rate) we find the answer is 52.5 mph.
+128408 We have that
Rate * Time = Distance
Call T the time we are looking for (in minutes)
So
45 ( T + 5) = 63 (T - 5) simplify
45T + 225 = 63T - 315
225 + 315 = T( 63 - 45)
540 = T ( 18)
540 / 18 = T = 30 minutes = 1/2 hr and 5 min = 1/12 hr
So...the distance =
45m/hr (1/2+ 1/12)hr = 45 ( 7/12) = 26.25 miles
So ... he needs to travel at Distance / Time (in hrs) = 26.25 / (1/2) = 52.5 mph to arrive on time
