An object is at rest at (2 ,1 ,6 )(2,1,6) and constantly accelerates at a rate of 1/4 m/s^214ms2 as it moves to point B. If point B is at (6 ,4 ,5 )(6,4,5), how long will it take for the object to reach point B? Assume that all coordinates are in meters.

2 Answers
Jul 20, 2017

The time is =6.39s=6.39s

Explanation:

The distance is

s=sqrt((6-2)^2+(4-1)^2+(5-6)^2)s=(62)2+(41)2+(56)2

=sqrt(16+9+1)=16+9+1

=sqrt26m=26m

The initial velocity is u=0ms^-1u=0ms1

The acceleration is =1/4ms^-2=14ms2

We apply the equation of motion

s=ut+1/2at^2s=ut+12at2

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

t=sqrt((2*sqrt26)/(1/4))t= 22614

t=6.39st=6.39s

Jul 20, 2017

6.3876.387 seconds

Explanation:

The distance between (2,1,6)(2,1,6) and 6,4,5)6,4,5) is

sqrt((6-2)^2+(4-1)^2+(5-6)^2)=sqrt(16+9+1)=sqrt26(62)2+(41)2+(56)2=16+9+1=26, and it is in meters.

As distance S=1/2at^2S=12at2, where aa is accelaration and tt is time taken, then

t=sqrt((2S)/a)=sqrt((2sqrt26)/(1/4))t=2Sa= 22614

= sqrt(8sqrt26)=2sqrt(2xx5.09902)=2sqrt10.19804=6.387826=22×5.09902=210.19804=6.387