What is the speed of an object that travels from #( -2,1,2) # to #( -3, 0,-6 ) # over #3 s#?

1 Answer
Dec 29, 2015

Answer:

#1.41"units""/s"#

Explanation:

To get the distance between 2 points in 3D space you effectively use Pythagoras in 2 D (x.y) and then apply that result to 3D (x,y,z).

Lets call #P=(-2,1,2)#

and #Q=(-3,0,6)#

Then #d(P,Q)=stackrel(rarr)(PQ)#

#=sqrt((-2+3)^2+(1-0)^2+(2-6)^2)#

#=sqrt(18)#

#=4.24#

#:.v=4.24/3=1.41"units/s"#