An object is at rest at #(6 ,5 ,9 )# and constantly accelerates at a rate of #4/3 m/s^2# as it moves to point B. If point B is at #(3 ,6 ,4 )#, how long will it take for the object to reach point B? Assume that all coordinates are in meters.

1 Answer
Shwetank Mauria
Jun 24, 2016

It will take #2.979# seconds.

Explanation:

The distance between two points #(x_1,y_1,z_1)# and #(x_2,y_2,z_2)# is given by

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

Hence distance between #(6,5,9)# and #(3,6,4)# is

#sqrt((3-6)^2+(6-5)^2+(4-9)^2)#

= #sqrt(3^2+1^2+5^2)=sqrt(9+1+25)=sqrt35=5.916#

(As distance covered is given by #S=ut+1/2at^2#, where #u# is initial velocity, #a# is accelaration and #t# is time taken. If body is at rest #S=1/2at^2# and hence #t=sqrt((2S)/a)#

As the coordinates are in meters, the time taken at an acceleration of #4/3# #m/sec^2# will be given by

#t=sqrt((2xx5.916)/(4/3))=sqrt((6xx5.916)/4)=sqrt(8.874)=2.979#

Related questions
See all questions in Acceleration
Impact of this question
1290 views around the world
You can reuse this answer
Creative Commons License