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.

## 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 SComment