An object is at rest at #(2 ,1 ,6 )# and constantly accelerates at a rate of #1/4 m/s^2# as it moves to point B. If point B is at #(6 ,4 ,5 )#, how long will it take for the object to reach point B? Assume that all coordinates are in meters.

Question

1425 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-20T15:03:43

The time is #=6.39s#

The distance is

#s=sqrt((6-2)^2+(4-1)^2+(5-6)^2)#

#=sqrt(16+9+1)#

#=sqrt26m#

The initial velocity is #u=0ms^-1#

The acceleration is #=1/4ms^-2#

We apply the equation of motion

#s=ut+1/2at^2#

#t=sqrt((2s)/a)#

#t=sqrt((2*sqrt26)/(1/4))#

#t=6.39s#

Answer 2 · 2017-07-20T15:06:44

#6.387# seconds

The distance between #(2,1,6)# and #6,4,5)# is

#sqrt((6-2)^2+(4-1)^2+(5-6)^2)=sqrt(16+9+1)=sqrt26#, and it is in meters.

As distance #S=1/2at^2#, where #a# is accelaration and #t# is time taken, then

#t=sqrt((2S)/a)=sqrt((2sqrt26)/(1/4))#

= #sqrt(8sqrt26)=2sqrt(2xx5.09902)=2sqrt10.19804=6.387#