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An object is at rest at #(4 ,1 ,2 )# and constantly accelerates at a rate of #7/5 m/s^2# as it moves to point B. If point B is at #(5 ,6 ,1 )#, how long will it take for the object to reach point B? Assume that all coordinates are in meters.

Answer 1

Leo Abeyta

The time is #=2.72s#

The distance #AB# is

#AB=sqrt((5-4)^2+(6-1)^2+(1-2)^2)=sqrt(1+25+1)=sqrt(27)m#

Apply the equation of motion

#s=ut+1/2at^2#

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

The acceleration is #a=7/5ms^-2#

Therefore,

#sqrt27=0+1/2*7/5*t^2#

#t^2=10/7*sqrt27#

#t=sqrt(10/7*sqrt26)=2.72s#

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Answer 2

William Audy

It will take 2 seconds for the object to reach point B.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.