meters wide). The thief runs on his side at 7 meters per second, you

run 9 meters per second. How fast are you approaching the thief if

A thief is 10 meters away (8 meters ahead of you, across a street 6

meters wide). The thief runs on his side at 7 meters per second, you

run 9 meters per second. How fast are you approaching the thief if

meters wide). The thief runs on his side at 7 meters per second, you

run 9 meters per second. How fast are you approaching the thief if

(a) you follow on your side of the street.

(b) you run toward the thief.

(c) you run away on your side of the street.

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Hi Lisa.

In this problem, we are asked to find the relative speed between you and the thief. This speed is the distance between you and the thief divided by time, so we need to find the distance between you and the thief, which itself depends on time. Since this is a 2-dimensional problem, set up an x-y-coordinate system with you initially at the origin. So your initial position is r_{1}=(0,0), the thief's initial position is r_{2}=(8,6), and your distance is Δr=√((8-0)²+(6-0)²)=10 (all numbers are in meters). The thief moves in the x-direction at 7 m/s, so his position as a function of time will be r_{2}=(8+7t,6).

(a) You move in the positive x-direction at 9 m/s. Your position at time t is r_{1}=(9t,0). Your relative distance is

Δr=√((8+7t-9t)²+(6-0)²)=√((8-2t)²+(6)²) = √(100-32t+4t²)

Notice that this distance is smallest at t=4s, when Δr=6m.

Your relative speed is v=Δr/t =√(100-32t+4t²) /t.

(c) I will do part (c) next, because it's in the same spirit as part (a).

You move in the negative x-direction at 9 m/s. Your position at time t is r_{1}=(-9t,0). Your relative distance is Δr=√((8+7t+9t)²+(6-0)²)=√((8+16t)²+(6)²) = √(100+256t+256t²)

Notice that this distance is smallest at t=0, when Δr=10m.

Your relative speed is v=Δr/t=√(100+256t+256t²) /t.

(b) "Running toward the thief" is a much harder problem, as you no longer follow a straight line path. Clearly, you are crossing the street, but never reach the other side of it. The tangent of the curve along which you are running is always pointing toward the thief. Such a curve is called a "tractrix" or "dog curve" (as in a dog running towards his owner, who moves along a straight line), and is quite complicated mathematically.

I will leave this part out, unless you really want to see it.

Hi Lisa;

I UPDATED MY ANSWER A FEW TIMES. I HOPE I DID NOT CONFUSE YOU. THIS IS MY FINAL ANSWER.

The issue of the 6 meter wide street is not included in this question.

Thief=7 m/s

You=9 m/s

distance=8 m

(9 m/s) - (7 m/s)=(8 m)/(x)

x=quantity of seconds

2=8/x

4=x

You need 4 seconds to catch up with him.

The distance is 8 meters in front of you, and 6 meters across the street. This is a triangle between you and the thief. We need to establish the length of the hypotenuse, and then establish how long it will take you to travel such distance, compared to that of the thief.

a^{2}+b^{2}=c^{2}

8^{2}+6^{2}=c2

64+36=c^{2}

100=c^{2}

10=c

As per the original question, he is 10 meters ahead.

10 meters=the distance you must run

distance=(speed)(time)

(distance)/(speed)=time

(10 m)/[(9 m/s)-(7 m/s)]=time

(10 m)/(2 m/s)=time

5 seconds=time

-(9 m/s + 7 m/s)=-16 m/s

The negative sign is because you are not approaching, you are reverse-approaching.

Hi Lisa;

I apologize for failing to notice that the thief is 8 meters ahead, rather than 10. The phrase "across the street" appears to contradict the issue of 10 meters away.

Hi Vivian and Lisa...I am reading this problem differently. For **a)**I see the original position of the you and the thief as a right triangle where the hypotenuse is 10 and the two legs are 6 and 8. After one second the two legs would be 6 and 6 and you would be the sq rte of 72 away or 8.28 meters. After 2 seconds the two legs would be 6 and 4 and you would be the sq rte of 52 away or 7.21 meters. After 3 seconds the two legs would be 6 and 2 and you would be the sq rte of 40 away or 6.32 meters away. After 4 seconds you would be directly across the street or 6 meters away. That means you are approaching at an average of 1 m/s. You would never get closer than 6 meters.

For part **c)** if you do the same thing as in **a)** the legs of the right triangle would start out 8 and 6 then 24 and 6, 40 and 6, 56 and 6 and 72 and 6 for the next 4 seconds. The hypotenuse would increase from 10 to 24.73 to 40.45 to 56.32 to 72.25 so you would have 62.25/4 or 15.56 m/s away from the thief. This would approach 16 m/s the longer you ran away from him. I haven't figured out part** b)** yet. Tell me what you think of my approach if you have time.

The speeds at issue are 7 and 9 m/s, and the distances at issue are 8 and 10 m. Therefore, you must catch-up with him. I therefore do not agree with your comment, "You would never get closer than 6 meters."

And while you are correct about your step-by-step analysis, it does not appear that the question is intended to be that complicated.

Furthermore, part (c) does not involve a triangle. You remain on your side of the street, turn around, and run.

Hi Vivian,

I may be thinking incorrectly but part **a)** says that the thief is running on his side of the street and you are running on your side which means the closest would be across the street from him after 4 seconds. You would still be 6 feet away. Part **c)** if you were always on different sides of the street then a line drawn from you to the thief would be the hypotenuse kitty corner across the street. One leg would always be 6 meters and the other leg would increase by 16 meters (7+9) every second. It starts at 8. This is an interesting problem.

Last word 2nd sentence above should have been meters not feet.

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## Comments

aandcbut I was doing it in one second increments instead of the correct way like you, expressing it as of function oftin general. The only thing I question in your equations is this. Shouldn't the initial 10m separation be taken into account when doing the relative average velocity?part ayour equation gives 6/4 = 1.5m/s away from the thief when it should really be -4/4 = 1m/s toward the thief. Expanding on your equation I believe it should beAvg v=(Δr-10)/t.c.Your equation yields anAvg vof 18.06239m/s away from the thief at 4 seconds but if you subtract off the initial 10m separation before you divide bytyou will get 15.56239m/s which makes more sense. As time goes on theAvg vwould increase but it could only approach (7m/s+9m/s) or 16m/s. Please tell me what you think. Dave S