An object is at rest at #(8 ,6 ,2 )# and constantly accelerates at a rate of #1/4 m/s^2# as it moves to point B. If point B is at #(2 ,8 ,2 )#, how long will it take for the object to reach point B? Assume that all coordinates are in meters.

1 Answer
Shwetank Mauria
Jul 1, 2016

It will take #7.11# 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 #(8,6,2)# and #(2,8,2)# is

#sqrt((2-8)^2+(8-6)^2+(2-2)^2)#

= #sqrt(6^2+2^2+0^2)=sqrt(36+4+0)=sqrt40#

(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 #1/4# #m/sec^2# will be given by

#t=sqrt((2xxsqrt40)/(1/4))=sqrt(4xx2xx6.3246)=sqrt(17.888)=7.11#

Related questions
See all questions in Acceleration
Impact of this question
1387 views around the world
You can reuse this answer
Creative Commons License