An object is at rest at #(8 ,3 ,8 )# and constantly accelerates at a rate of #7/4 m/s^2# as it moves to point B. If point B is at #(6 ,3 ,4 )#, how long will it take for the object to reach point B? Assume that all coordinates are in meters.

1 Answer
Mar 18, 2017

Answer:

The time is #=2.26s#

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=7/4ms^-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=(8,3,8)# and #B=(6,3,4)#

So,

#s=sqrt((6-8)^2+(3-3)^2+(4-8)^2)#

#=sqrt(4+0+16)#

#=sqrt20#

From the equation of motion,

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

#t^2=(2*sqrt20)/(7/4)=5.11#

#t=sqrt5.11=2.26s#