Question 639611: Some bank robbers leave town, speeding at 71 mph. Nine minutes later, the police give chase, traveling at 80 mph. How long will it take after the robbery for the police to overtake the robbers? (Round your answer to three decimal places.)
I have tried over and over to get this right. This is what I have tried:
I made a chart for the robbers saying their rate is 71 mph and time is t+.15 (9 minute head start). Then I made one for cops saying their rate is 80 mph and their time is t. Therefore, the distance of robbers are 71(t+.15) and the distance of cops are 80t. Then I made the equation look like this:
80t = 71(t+.15)
80t = 71t + 10.65 (distribution)
9t = 10.65
t = 1.183
Please help. What am I doing wrong? :(
Answer by DrBeeee(684)   (Show Source):
You can put this solution on YOUR website! You are using the wrong expression for time.
Let t1 = the time of the robbers, and
t2 = time of the cops.
As you correctly have that they both travel the same distance, d.
The robbers travel at 71 mph, their distance is
(1) d = 71*t1 and the cops travel
(2) d = 80*t2.
Now what is the relationship between t1 and t2? Since the robbers started first we will start the "chase" at t1 = 0, or the time of the robbery.
Then when t1 is 9 minutes or 9/60 hours, t2 starts, therefore
(3) t2 = t1 - 9/60
Substituting (3) into (2) yields
(4) d = 80*(t1 - 9/60)
Now equate (1) and (4) yields
(5) 71*t1 = 80*(t1 - 9/60)
Simplifying (5) gives
(6) 9*t1 = 80*(9/60) or
(7) t1 = 4/3 hr
Let's check this.
The robbers travelled
d = 71*(4/3) miles.
The cops travelled
d = 80*(4/3 - 9/60)
d = 80[(240-27)/180]
d = 80[213/180]
d = 80[(3*71/(3*60)]
d = 71*(80/60) or
d = 71*(4/3)
which is the same as the robbers.
My answer (the correct time) is
the time that the cops apprehend the robbers is 4/3 hour AFTER the robbery took place and robbers sped off.
In your solution time is starting at 9 minutes (after the robbery), not zero. To get the correct total time add 9/60 to your 71/60 and get 80/60. This is the time AFTER the robber.
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