A policeman ran after a thief 1440 km ahead. If the speed of the thief is 110 kilometers per hour and the speed of the police is 128 kilometers per hour, how much will the police run and catch the thief?
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So if we just think about motion, distance traveled is the product of velocity and time. For the cop, lets call the distance he’s traveled Dc and since he runs 1km in 8 minutes his velocity is (1km/8m)*t where t is time in minutes. So his distance traveled as a function of time is the linear equation:
Dc = (1km/8m)*t
The thief we can describe in the same way, except since he got a 100m head start we can add .1km to the linear equation
Dt= (1km/10m)*t + .1km
Now in order for the cop to catch the thief, they will both have to have traveled the same distance within the same time frame. In otherwords, we know that at some value of t, Dc=Dt and that is the point at which the cop has caught up to the thief.
You can do this one of two ways, graphically, you’ll notice they both are linear y=mx+b equations, so if you graph them both with distance being y, the point at which they intersect is the point at which the thief is caught and you will have the value for both the time it took and the distance that was traveled.
Or you can use the substitution method, since we want to know the details about where Dc=Dt, setting both right hand sides of the equations equal to each other, solving for t, and then plugging that back into either cop or theif’s equation will give you the distance traveled.
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