An object is at rest at (4 ,8 ,2 ) and constantly accelerates at a rate of 4/3 m/s^2 as it moves to point B. If point B is at (8 ,1 ,7 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.

1 Answer
Apr 19, 2017

The time is =3.77s

Explanation:

The distance between the points A=(x_A,y_A,z_A) and the point B=(x_B,y_B,z_B) is

AB=sqrt((x_B-x_A)^2+(y_B-y_A)^2+(z_B-z_A)^2)

d=AB= sqrt((8-4)^2+(1-8)^2+(7-2)^2)

=sqrt(4^2+(-7)^2+5^2)

=sqrt(16+49+25)

=sqrt90

We apply the equation of motion

d=ut+1/2at^2

u=0

so,

d=1/2at^2

t^2=(2d)/a=(2sqrt90)/(4/3)

=14.23

t=sqrt(14.23)=3.77s