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.

2 Answers
Jul 20, 2017

Answer:

The time is #=6.39s#

Explanation:

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#

Jul 20, 2017

Answer:

#6.387# seconds

Explanation:

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#