Question

At noon, ship A is 100km west of ship B. Ship A is sailing east at 30km/h and ship B is sailing north at 40km/h. How fast
is the distance between the ships chaning at 4pm?

Answer 1

At noon, ship A is 100km west of ship B. Ship A is sailing east at 30km/h and ship B is sailing north at 40km/h. How fast is the distance between the ships changing at 4pm?

Solution:

Let at noon ship A was at the point A and ship B was at the point B. After $t$ hours Ship A is at the point $A_{1}$ and ship B at the point $B_{1}$

So after $t$ hours the distance between the ships will be:

$$S = A_{1}B_{1} = \sqrt{BA_{1}^{2} + BB_{1}^{2}}$$ $$BA_{1} = AB - AA_{1}$$ $$AB = 100 \text{ km} - \text{given}$$ $$AA_{1} = v_{A} \cdot t$$ $$v_{A} = 30 \text{ km/h} - \text{given}$$ $$\text{So } BA_{1} = 100 - 30t$$ $$BB_{1} = v_{B} \cdot t$$ $$v_{B} = 40 \text{ km/h} - \text{given}$$ $$BB_{1} = 40 \cdot t$$ $$\text{So } S = \sqrt{(100 - 30t)^{2} + (40t)^{2}}$$

The rate of change the distance is:

$$V = \frac{dS}{dt} = \frac{-60t + 80t}{2\sqrt{(100 - 30t)^{2} + (40t)^{2}}} = \frac{10t}{\sqrt{(100 - 30t)^{2} + (40t)^{2}}}$$

When $t = 4$

$$V = \frac{10 \cdot 4}{\sqrt{(100 - 30 \cdot 4)^{2} + (40 \cdot 4)^{2}}} = 0.25 \text{ km/h}$$