Question 152447: help me pleasee
a highway patrolman spots a speeding car. he clocks it at 70 mph and takes after it .05 mile behind. if the patrolman travels at an average rate of 90 mph, how long before he over takes he car?
Found 2 solutions by nerdybill, bucky:
Answer by nerdybill(7384)   (Show Source):
You can put this solution on YOUR website! a highway patrolman spots a speeding car. he clocks it at 70 mph and takes after it .05 mile behind. if the patrolman travels at an average rate of 90 mph, how long before he over takes he car?
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You'll need to apply the "distance formula":
d = rt
where
d is distance
r is rate or speed
t is time
.
In general:
"distance traveled by patrolman" = "distance behind" + "distance traveled by speeder"
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Let t = time it takes for patrolman to catch up
.
90t = .05 + 70t
20t = .05
t = .05/20
t = 0.0025 hours
.
Or, in terms of minutes:
0.0025 * 60 = .15 minutes
Or, in terms of seconds:
.15 * 60 = 9 seconds
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Note: check your original problem. Are you sure the patrolman was ONLY behind .05 miles and not .5 miles? Because .05 mile is only 264 feet away.
Answer by bucky(2189)  (Show Source):
You can put this solution on YOUR website! To solve this problem you can use the equation that Distance = speed * time.
.
The speeding car and the patrolman will both have traveled the same distance when the patrolman
catches up to the speeder.
.
Let t represent the time that it takes the patrolman to catch up to the speeding car. The distance
that the patrolman will travel is his speed times the time it takes him to catch up and this
distance can be represented by the equation:
.
D = 90 * t
.
Meanwhile, the speeding car has gone 0.05 miles before the patrolman starts out. So the
distance that the speeding car travels is 0.05 miles before the clock starts. Then the
speeding car continues to travel at 70 mph for the time t. In equation form this becomes:
.
D = 0.05 + (70*t)
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But as indicated above, the two distances traveled are equal. So setting the right side
of these two equations equal results in:
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90*t = 0.05 + 70*t
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Subtracting 70t from both sides reduces this equation to:
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20t = 0.05
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Finally, solve for t by dividing both sides by 20 to get:
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t = 0.05/20 = 0.0025
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This tells you the time it takes the patrolman to catch the speeder is 0.0025 of an hour
and since an hour is 60 minutes, the time in minutes to catch the speeder is:
.
t = 0.0025 * 60 = 0.15 minutes
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and since a minute is 60 seconds, the time to catch the speeder is:
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t = 0.15 * 60 = 9 seconds
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This is very, very little time to reach an average speed of 90 mph if the patrol car
was stopped by the side of the road when the speeder went by.
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This would lead you to assume that patrolman was going at 70 mph behind the speeder when
he clocked him, and from that speed the patrolman sped up to the average speed of 90 mph.
.
Just make sure that the patrol car didn't start out a half-mile (0.5 mile) behind the speeding
car. If that were the case the equation for the speeding car would be:
.
D = 0.5 + 70t
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and the equality would be:
.
90t = 0.5 + 70t
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Subtracting 70t from both sides results in:
.
20t = 0.5
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and solving for t by dividing both sides by 20 gives:
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t = 0.5/20 = 0.025 hr
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which translates to:
.
0.025 * 60 = 1.5 minutes
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Without analyzing the physics of the problem, this would seem slightly more reasonable time
for "real-world" performance of a patrol car that accelerates from stop to attain an
average speed of 90 mph.
.
Hope this helps you to understand the problem.
.
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