a submarine is observed to be at the point (-1,1,-1/4) at t=0 and 45 minutes later it is observed to be at the point (2,4,-1/5). here the water surface is positioned in the xy-plane and z vertically up and with the unit in nautical miles.
Suppose now the submarine goes in a straight path with constant speed.
A: What´s the submarines velocity v and its speed [MATH]\mid V \mid [/MATH].
B: When and where does the submarine reach the surface?
For A my thinking was that [MATH]\Delta \vec{r}=r_1-r_0 \iff \Delta \vec{r} = (3,3,1/20)[/MATH], note that [MATH]r_1; (2,4,-1/5)[/MATH] and [MATH]r_0: (-1,1,-1/4)[/MATH]since Velocity=displacement/time I then have that [MATH]\vec{V}=\dfrac{\Delta r}{t}=\dfrac{\Delta r}{46 \cdot 60}=(1/900,1/900,1/5400)[/MATH]and so [MATH]\mid V \mid = \sqrt{\vec{V}.\vec{V}}=0,00157...[/MATH]
Could that be correct? It feels like the number is too small.
Suppose now the submarine goes in a straight path with constant speed.
A: What´s the submarines velocity v and its speed [MATH]\mid V \mid [/MATH].
B: When and where does the submarine reach the surface?
For A my thinking was that [MATH]\Delta \vec{r}=r_1-r_0 \iff \Delta \vec{r} = (3,3,1/20)[/MATH], note that [MATH]r_1; (2,4,-1/5)[/MATH] and [MATH]r_0: (-1,1,-1/4)[/MATH]since Velocity=displacement/time I then have that [MATH]\vec{V}=\dfrac{\Delta r}{t}=\dfrac{\Delta r}{46 \cdot 60}=(1/900,1/900,1/5400)[/MATH]and so [MATH]\mid V \mid = \sqrt{\vec{V}.\vec{V}}=0,00157...[/MATH]
Could that be correct? It feels like the number is too small.