Question 955979
Michael drove to work on Monday at 45 mph and arrived one minute early. The employee drove to work on Tuesday, leaving home at the same time driving 40 mph and arriving one minute late. How far does Michael live from work? What speed does he need to drive to arrive 5 minutes early?
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min=1/60 hr
d=distance
s=speed
let x=0n-time travel time
distance/speed=travel time
d/45=x-(1/60)
d/40=x+(1/60)
45(x-(1/60)=40(x+(1/60)
45x-45/60=40x+40/60
5x=95/60
x=95/300 hr or 19 min (on-time travel time)
d=45(x-(1/60)=45(95/300-5/300)=45(90/300)=13.5 mi
d/s=95/300-5/60
13.5/s=95/300-25/300=70/300
s=13.5/(70/300)≈57.86 mph
How far does Michael live from work? 13.5 mi
What speed does he need to drive to arrive 5 minutes early? 57.86 mph