An object is at rest at #(8 ,2 ,5 )# and constantly accelerates at a rate of #7/4 m/s^2# as it moves to point B. If point B is at #(2 ,3 ,7 )#, how long will it take for the object to reach point B? Assume that all coordinates are in meters.

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Answer 1 · 2017-03-18T15:03:20

The time taken is #=2.71s#

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=(8,2,5)# and #B=(2,3,7)#

So,

#s=sqrt((2-8)^2+(3-2)^2+(7-5)^2)#

#=sqrt(36+1+4)#

#=sqrt41#

From the equation of motion,

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

#t^2=(2*sqrt41)/(7/4)=7.32#

#t=sqrt7.32=2.71s#