An object is at rest at #(6 ,5 ,9 )# and constantly accelerates at a rate of #4/3 m/s^2# as it moves to point B. If point B is at #(3 ,6 ,4 )#, how long will it take for the object to reach point B? Assume that all coordinates are in meters.

Question

An object is at rest at #(6 ,5   ,9    )# and constantly accelerates at a rate of #4/3  m/s^2# as it moves to point B.  If point B is at #(3  ,6   ,4  )#, how long will it take for the object to reach point B? Assume that all coordinates are in meters.

Olivia Askew · Accepted Answer

It will take #2.979# seconds.

The distance between two points #(x_1,y_1,z_1)# and #(x_2,y_2,z_2)# is given by #sqrt((x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2)# Hence distance between #(6,5,9)# and #(3,6,4)# is #sqrt((3-6)^2+(6-5)^2+(4-9)^2)# = #sqrt(3^2+1^2+5^2)=sqrt(9+1+25)=sqrt35=5.916# (As distance covered is given by #S=ut+1/2at^2#, where #u# is initial velocity, #a# is accelaration and #t# is time taken. If body is at rest #S=1/2at^2# and hence #t=sqrt((2S)/a)# As the coordinates are in meters, the time taken at an acceleration of #4/3# #m/sec^2# will be given by #t=sqrt((2xx5.916)/(4/3))=sqrt((6xx5.916)/4)=sqrt(8.874)=2.979#

Carson Alexander · Answer

undefined Use the kinematic equation: $ s = ut + \frac{1}{2}at^2 $  
where $ s $ is the displacement, $ u $ is the initial velocity, $ a $ is the acceleration, and $ t $ is the time.  
Rearrange the equation to solve for $ t $: $ t = \sqrt{\frac{2s}{a}} $  
Calculate the displacement ($ s $) using the distance formula.  
Initial position (A): $ (6, 5, 9) $  
Final position (B): $ (3, 6, 4) $  
Substitute values into the equation and solve for $ t $.  
The time it takes for the object to reach point B is approximately 2.45 seconds.