An object is at rest at #(4 ,7 ,1 )# and constantly accelerates at a rate of #4/3 m/s^2# as it moves to point B. If point B is at #(3 ,1 ,6 )#, how long will it take for the object to reach point B? Assume that all coordinates are in meters.

2 Answers
Mar 19, 2016

Answer:

#t=3,44" s"#

Explanation:

#"distance between two points:"#
#Delta x=3-4=-1#
#Delta y=1-7=-6#
#Delta z=6-1=5#
#s=sqrt((Delta x)^2+(Delta y)^2+(Delta z)^2)#
#s=sqrt((-1)^2+(-6)^2+5^2)#
#s=sqrt(1+36+25)" "s=sqrt62" m"#
#s=7,87 " m"#

#s=1/2*a t^2#
#7,87=1/cancel(2)*cancel(4)/3*t^2#
#t^2=(7,87*3)/2#
#t^2=11,81#
#t=sqrt (11,81)#
#t=±3,44#
#t=cancel(-3,44)" there is no negative time"#
#t=3,44" s"#

Mar 19, 2016

Answer:

#t=sqrt(3sqrt(31/2))=3.4367#

Explanation:

Given Location of an object,
Rest position #A(4,7,1)#
New Location #B(3,1,6)#
Acceleration, #a=4/3m/s^2#
Find the time it take to move from #A -> B#
Using the distance formula we find the distance from #A->B#
#bar(AB) =r= sqrt((4-3)^2+(7-1)^2 +(1-6)^2) = sqrt(62)#
Now we need to use the kinematic equation of motion that uses
#t, a, r# as variable.
#F(r,t;v_o,a) => r = v_ot +1/2at^2 #; #v_o = 0, r=sqrt(62); a=4/3m/s^2#
#sqrt(62) = 1/2*4/3t^2#
#t=sqrt(3sqrt(31/2))=3.4367#