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An object is at rest at #(9 ,7 ,5 )# and constantly accelerates at a rate of #5/4 m/s^2# as it moves to point B. If point B is at #(1 ,9 ,2 )#, 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 #=3.75s#

The distance #AB# is

#AB=sqrt((1-9)^2+(9-7)^2+(2-5)^2)=sqrt(64+4+9)=sqrt(77)m#

Apply the equation of motion

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

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

The acceleration is #a=5/4ms^-2#

Therefore,

#sqrt77=0+1/2*5/4*t^2#

#t^2=8/5*sqrt77#

#t=sqrt(8/5*sqrt77)=3.75s#

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Answer 2

Layla Ahmed

It will take approximately 4.08 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.