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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?

 Jan 5, 2021
 #1
avatar+292 
+1

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.

 Jan 5, 2021
 #2
avatar+114361 
+2

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  

 

 

cool cool cool

 Jan 5, 2021

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