An object is at rest at #(2 ,1 ,5 )# and constantly accelerates at a rate of #3 m/s# as it moves to point B. If point B is at #(6 ,7 ,5 )#, 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 #2.193# 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 #(2,1,5)# and #(6,7,5)# is

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

= #sqrt(4^2+6^2+0^2)=sqrt(16+36+0)=sqrt52=2sqrt13#

(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 #3# #m/sec^2# will be given by

#t=sqrt((2xx2sqrt13)/3)=sqrt((4xx3.606)/3)=sqrt(4.808)=2.193#

Related questions
See all questions in Acceleration
Impact of this question
1356 views around the world
You can reuse this answer
Creative Commons License