Math word problem involving miles, speed, and time.

Question

The subway train offers access to all 12 intersections downtown. If every intersection is .75 miles apart and the train travels at 20 mph and spends 45 seconds at every stop to unload and onboard new customers, how long will it take to get from the first intersection to the last intersection? 

Answer 1

First, a couple of assumptions:

1. We ignore the consideration of how fast the train accelerates to 20 mph and how quickly its doors open

2. The clock starts when the doors open at station 1 (initiating its load/unload cycle of 45 seconds.)

 

The time required for the train to load/unload at station x and travel to station x+1 would then be the 45 seconds for the load/unload cycle and .75 miles / (20 miles/hour) = .0375 hours * (3600 sec/hr) = 145 sec

 

Thus, it takes 190 sec for a train just arrived at station x to arrive at station x+1. Since there are 11 station transits to arrive at station 12, the total time is 11*190 seconds or just shy of 34 minutes.

Answer 2

The train starts at station 1. The train stays at each station for 45 seconds. The question asks how long it will take to get from the first station to the last station. The train stays at each station for 45 seconds, but when the train gets to the last station, the problem ends; therefore we don't have to consider the 45 seconds it stays at station 12. Again, the problem ends when the train gets to station 12 !

We have to consider 45 seconds at each station for only 11 stations; therefore we have (11*45)/3600. We divide by 3600 seconds in an hour.

distance=rate*time and time=distance/rate

To go from station 1 to station 12, the train moves only 11 times.

(0.75miles/20 mph)*11

(11*45)/3600=495/3600=0.1375 hours

0.75/20=0.0375

0.0375*11=0.4125 hours

0.1375+0.4125=0.55 hours

0.55*60(minutes)=33 minutes to get from station 1 to station 12

Remember, the problem asks how long it takes to get from station 1 to station 12. When the train gets to station 12, the problem ends.

Answer 3

I created an equation that is: (45/3600)+((0.75/20)+(45/3600))*12 ; the first term is for the first delay without velocity or movement and the following address the 11 other stations or 12 spaces.

This will give me ; total time= 27.75 minutes = 28 minutes.