Question 977637: two thieves rob a jewelry store and make their getaway at 10 am, traveling at 85 mph. at noon, the police leave the store in hot pursuit of the robbers. if the police travel 105 mph, how long will it take them to catch the robbers? at what time of day does this occur? how far from the jewelry store is the arrest made?
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! d = distance traveled in miles
t = time (in hours) that has elapsed since robbers started their getaway
t = 0 represents 10 AM
Two hours later, t = 2 represents 12 PM noon (same day) which is when the cops start from the store and chase.
The robbers travel a distance of d = 85t. The cops travel a distance of d = 105(t-2). The "t-2" represents the idea that the robbers get a 2 hr head start. So you have to subtract 2 hrs from the cop's time. Notice how if t = 2 for the cops, then d = 105(t-2) = 0. That ensures the cops have d = 0 when it is noon.
System of Equations:
d = 85t
d = 105(t-2)
The goal is to find d & t. Start with one equation, then plug in the other (substitution) to solve for t.
d = 85t
105(t-2) = 85t
105t-210 = 85t
105t-210+210 = 85t+210
105t = 85t+210
105t-85t = 85t+210-85t
20t = 210
t = 210/20
t = 10.5
It takes 10.5 hours for the cops to catch the robbers. The cops catch the robbers at 10:30 pm on the same day (when the cops started at noon).
Use this to find d.
d = 85t
d = 85*10.5
d = 892.5
The cops travel 892.5 miles. The robbers travel the same distance.
The arrest is made 892.5 miles away from the store.
Notes: The following assumptions were made
1) The speeds are held constant
2) The road is completely straight
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