Hi I\'m trying to solve this problem and I don\'t think I\'m doing anything wron

Question

Hi

I'm trying to solve this problem and I don't think I'm doing anything wrong but it still says the answer is incorrect.

Can somebody please help me?

Jessica and Matthew are running toward the point P along the straight paths that make a fixed angle of (Figure 1). Suppose that Matthew runs with velocity va (m/s) and Jessica with velocity vb (m/s) Let f(x, g) be the distance from Matthew to Jessica when Matthew is z meters from P and Jessica is y meters from P FIGURE 1 . show that f(z, y) = Atv'-2zycose ·Assume that -/3. Use the Chain Rule to determine the rate at which the distance between Matthew and Jessica is changing when z = 10 y = 11, va-2 m/s, and vb = 3 m/s df -| 2.56

Explanation / Answer

The hard part of this problem is we don't know what t is

Since va = 2m/s ,so AM = 2t
and BN = 3t.

Let t=10 ( after 10 sec )

2(10) +10= 30
So AP should be 30m , so that when the guy runs after 10 sec,
the distance PM is 10m
3(10) +11= 41
Similarly, PN = 41 -3t

After 10 sec, PN is 41- 10(3) = 41 - 30 =11m

Now MN^2 = PM^2 +PN^2 - 2PM.PN cos (pi/3)

Plug in the expression

MN^2 = ( 30 - 2t )^2 + ( 41 - 3t )^2 - 2 ( 10 ) (11) ( 1/2)

MN^2 = 13t^2-366t+2471

MN = sqrt ( 13t^2-366t+2471 )

Take d(MN)/dt with respect to t

==> 13t-183 / sqrt{13t^2-366t+2471}

then plug in t= 10 ==> answer = -5.03054 m/s

It's interesting to note that you can choose another different t
and it still gives the same answer.
for example t = 5sec. Of course, the expression of PA and PB must be different.

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