An object is at rest at #(1 ,5 ,1 )# and constantly accelerates at a rate of #4/3 m/s^2# as it moves to point B. If point B is at #(9 ,4 ,8 )#, how long will it take for the object to reach point B? Assume that all coordinates are in meters.

1 Answer
Apr 16, 2016

Answer:

#t~=4" s"#

Explanation:

#A=(1,5,1)" the point of A"#

#B=(9,4,8)" the point of B"#

#A_x=1" "A_y=5" "A_z=1#

#B_x=9" "B_y=4" "B_z=8#

#"let's calculate distance between the points A and B"#

#Delta s=sqrt((B_x-A_x)^2+(B_y-A_y)^2+(B_z-A_z)^2)#

#Delta s=sqrt((9-1)^2+(4-5)^2+(8-1)^2)#

#Delta s=sqrt(8^2+(-1)^2+7^2)#

#Delta s=sqrt(64+1+49)#

#Delta s=10,68" m"#

#a=4/3 m/s^2 "acceleration of object"#

#t="elapsed time from the point of A to the point of B"#

#Delta s=1/2 *a*t^2#
#"distance equation for an object moving from rest"#

#10,68=1/2*4/3*t^2#

#t^2=(3*10,68)/2#

#t^2=16,02#

#t~=4" s"#