I'm stuck!...A highway patrolman traveling @ the speed limit is passed by a car going 15mph faster than he is.

Question

After one minute, the patrolman speeds up to 100 mph. How many minutes after speeding up until the patrolman catches up with the speeding car. The speed limit is 55 mph.

I thought I knew how to solve this problem, but I can't seem to get the right answer. This is what I did, please tell me where I went wrong/how to solve it properly

distance= speed * time
the ? ask when the hp(A) will catch up w/ the car(B)
in other words when will d(A)=d(B)
d(A)=100((x-1)/60)
{ A is 1 min. behind; I put this over 60 bc I need x in min. }
d(B)=70(x/60)
so...
100((x-1)/60) = 70(x/60) and solve for x
(10x-10)/6 = (7x)/6
60x-60 = 42x
18x=60
x=60/18 = 3.33 minutes
...I've been over and over this problem & I need some help...thanks so much!

Answer 1

The patrol car is traveling at two different speeds so there should be two terms in the equation for d(A).

d(A) = (100x/60) - 15/60 (distance lost in the first minute 70 mph minus 55 mph is 70/60 - 55/60 = 15/60)

Your d(B) is correct.

So

(100x/60) - 15/60 = 70x/60

100x -15 = 70x

30x =15

I get x = 15/30 = .5 minutes

(of course, this question is faulty, because cars can't go from 55 MPH to 100 MPH instantaneously, and we're not told how long it takes to speed up, so it would actually take a little longer, but whatever...)

(**I did edit my answer, I saw that I missed something after reading bipabeop's answer, but I did write the equation in your format. **)

Answer 2

does the patrol guy just accelerate up to 100mph in an instant or does he gradually accelerate. If its assumed to be an instant;

The speeding car will travel a distance of 70 x (1/60) = 1.167 miles (speed x time) and at the same time the patrol guy will travel 55 x (1/60) = 0.917

This mean while the patrol guy was going slower the speeder gains 0.25 of a mile.

Now the patrol car speeds up to 100mph in an instant and the speeding car stays at 70 mph. The relative speed of the cop car is now 30 mph (100- 70).

So the time taken for the patrol car to catch the speeding car is distance divided by speed.

0.25/30 = (1/120) hours which is 30 seconds

Answer 3

easy...... 15/60=.25= 1/4 mile in 1 minute.......this is the speedster
30/60=.5 = 1/2 mile in 1 minute.......this is the copper
it will take the copper 30 seconds alone to reach the point of where the speedster was when the copper hit the pedal, but at this point, the speedster is now 1/8 of a mile ahead of him,which will take the copper 15 seconds to do that same 1/8 mile. the speedster will go another 1/8 mile before the copper catches up.meaning 15 more seconds to the coppers time.thus meaning that 30+15+15=1 whole minute.yup, takes the copper 1 minute to catch up. And now, I too have a headache. hope this makes since to you.

Answer 4

Cop is going 55
Speeder going 70 = cop + 15
60 MPH=1 mile per minute (60 Minutes in an hour)
Speeder gains one quarter (.25) mile in one minute 15/60
Cop accelerates to 100
That is speeder (70) plus 30 (30/60)
Cop regains the quarter mile lost in half a minute.

Answer 5

you at the instant are not allowed to hurry on a 4 lane hwy - even to bypass. the only time they provide you a buffer for rushing while passing is on a 2 lane highway, so as which you will thoroughly get lower back into your lane in a lifelike quantity of time. you would be able to desire to attempt to dispute the value ticket in court docket, yet maximum probable you will lose. Sorry

Answer 6

holy chit......u just gave me a headache!