An object is at rest at #(4 ,5 ,6 )# and constantly accelerates at a rate of #5/3 m/s^2# as it moves to point B. If point B is at #(7 ,5 ,3 )#, how long will it take for the object to reach point B? Assume that all coordinates are in meters.

1 Answer
Oct 25, 2017

Answer:

#" Appro. "2.26 sec.#

Explanation:

Suppose that, the object is at rest the point (pt.) #A=A(4,5,6).#

Its final position is at the pt. #B=B(7,5,3).#

Hence, using the distance (dist.) formula, the dist. travelled by

the object, i.e.,

#"Dist. "AB=sqrt{(7-4)^2+(5-5)^2+(3-6)^2}, or,#

#AB=s==sqrt18.#

Recall that, #s=ut+1/2at^2........(star),#

where,

#s"=dist., "u"=initial velocity, "t"=time, and, "a"=accelaration."#

In our case, #s=sqrt18, u=0, a=5/3.# Sub.ing in #(star),# we get,

#sqrt18=0+1/2*5/3t^2 :. t^2=(6*3sqrt2)/5=(3.6)(1.414),#

# t^2~~5.0912,#

# rArr t~~sqrt5.0912~~2.26 sec.#