An object is at rest at #(4 ,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 ,2 ,6 )#, how long will it take for the object to reach point B? Assume that all coordinates are in meters.

1 Answer
Jun 18, 2016

Answer:

It takes #3.14# s.

Explanation:

The equation of motion for a constant accelerating object is

#s=1/2at^2# where #s# is the space traveled, #a# is the acceleration and #t# is the time.

We are interested in the time that is

#t=sqrt((2s)/a)#.

The acceleration is given, then we need the space.
The distance between two point in space #p_1(x_1, y_1, z_1), p_2(x_2, y_2, z_2)# is given by

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

For us

#d=sqrt((4-7)^2+(5-2)^2+(1-6)^2)#

#=sqrt((-3)^2+(3)^2+(-5)^2)#

#=sqrt(9+9+25)=sqrt(43)\approx6.56#.

This is the distance that the object has to travel, so we can write
#s=6.56# m. We now have all the ingredients

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

#=sqrt((2*6.56" m")/(4/3"m"/"s"^2))#

#=sqrt((13.12" m")/("m")*3/4 "s"^2)#

#=sqrt(9.84" s"^2)#

#=sqrt(9.84" s"^2)\approx3.14" s"#.