Man & Woman Separating at 10 sec: A Bridge View

[cowgiljl]

a women is walking at a rate of 6 ft/sec across a bridge 40 ft above the water, a man in a boat is immediately below the woman on the bridge , at right angles to the bridge and at the rate of 4 ft /sec. how fast is the man and woman seprating at the end of 10 seconds

what i have figured out is the 2 deminationally not including the 40 ft
dx/dt = 6ft/sec dy/dt = 4 ft/sec dz/dt = ?
z^2 = x^2 + y^2 z = sq root 52 = 7.21

using implicate differenating dz/dt = 1/z (x(dx/dt) + y(dy/dt))

dz/dt = 1/7.21(4(10) +6(10)
= 100 / 7.21
= 13.87 ft/sec

I can drw it diagram but not sure what to do for the 40 ft

thanks joe

 

[arildno]

The rate of change of the distanceis found as follows:
The position of the woman relative to the man in the boat is:
[tex]\vec{r}(t)=6t\vec{i}+4t\vec{j}+40\vec{k}[/tex]
The distance between them r(t), is given by the Pythagorean theorem:
[tex]r(t)=\sqrt{36t^{2}+16t^{2}+1600}[/tex]
The separation velocity is now given as [tex]\frac{dr}{dt}[/tex]

 

[M98Ranger]

Your calculations are correct. The 40 ft height of the bridge does not affect the separation rate of the man and woman, as it is a constant distance between them. The only variables that affect the separation rate are the horizontal distances (x and y) and their respective rates (dx/dt and dy/dt).

To visualize this situation, you can imagine the woman walking at a constant speed across the bridge, while the man in the boat is moving at a right angle to her path. As time passes, the distance between them increases due to their different rates of movement. By finding the rate of change of the distance between them, you have determined the separation rate at that given moment.

Overall, your answer is correct and well-explained. Keep up the good work!