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.

Question

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.

Isabella Alvarez · Accepted Answer

It will take #2.193# seconds.

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#

Isabella Alvarez · Answer

undefined To calculate the time it takes for the object to reach point B, you can use the kinematic equation:

$$ s = ut + \frac{1}{2}at^2 $$

where:
- $ s $ is the displacement (distance between the initial and final positions),
- $ u $ is the initial velocity (which is 0 since the object is at rest),
- $ a $ is the acceleration,
- $ t $ is the time.

Rearrange the equation to solve for $ t $:

$$ t = \sqrt{\frac{2s}{a}} $$

Substitute the values:

$$ t = \sqrt{\frac{2 \sqrt{(6-2)^2 + (7-1)^2 + (5-5)^2}}{3}} $$