An object is at rest at #(3 ,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 #(7 ,9 ,2 )#, how long will it take for the object to reach point B? Assume that all coordinates are in meters.

1 Answer
Mar 17, 2017

Answer:

The time taken is #=2.94s#

Explanation:

We are going to apply the following equation of motion :

#s=u_0t+1/2at^2#

#u_0=0#

So,

#s=1/2at^2#

#a=4/3ms^-2#

The distance between 2 points #A=(x_1,y_1,z_1)# and #B=(x_2.y_2.z_2)# is

#=sqrt((x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2)#

Here,

#A=(3,5,1)# and #B=(7,9,2)#

So,

#s=sqrt((7-3)^2+(9-5)^2+(2-1)^2)#

#=sqrt(16+16+1)#

#=sqrt33#

From the equation of motion,

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

#t^2=(2*sqrt33)/(4/3)=8.62#

#t=sqrt8.62=2.94s#