# Math Help - Question Difficulty

Thanks!

2. ## nr. 6 and nr. 7 missing

Hi,

to 1) The fishing boat needs $\frac{400}{8.5}\approx 47\ s$
the ferry needs $\frac{120}{3}= 40\ s$

So the ferry is 21 m south of the fishing boat when it reaches the point X. (Quite a narrow escape!)

to 2) Use the Pythagoran rule: $d=\sqrt{400^2+120^2}\approx 417.6 m$

to 3) Use the Pythagoran rule. One leg is 400-8.5*t and the other is 120-3*t. So you get:
$d(t)=\sqrt{(400-8.5t)^2+(120-3t)^2}$

to 4) You are looking for the minimum distance. So d(t) has a minimum if D(t)=(d(t))² has a maximum. For convenience I use D(t)
$D(t)=81.25t^2-7520t+174400$
$\frac{d(D(t))}{dt}=162.5t-7520$. Thus t ≈ 46.3 s.

That means 0.7 s before the fishing boat arrives at point X the vessels are closest to each other.

d(46.3) ≈ 19.97 m ≈ 20 m

to 5) The visibility is 50 m. So the distance is less or equal 50 m.

$d(t)=\sqrt{(400-8.5t)^2+(120-3t)^2}=50$. Solve for t. You'll get 2 values:

t_1 = 41.2 s or t_2 = 51.4 s

During these approximately 10 s they are visible to each other.

EB