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An object is at rest at #(8 ,9 ,2 )# and constantly accelerates at a rate of #4/3 m/s# as it moves to point B. If point B is at #(3 ,1 ,6 )#, 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 #=3.92s#

The distance #AB# is

#AB=sqrt((3-8)^2+(1-9)^2+(6-2)^2)=sqrt(25+64+16)=sqrt(105)m#

Apply the equation of motion

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

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

The acceleration is #a=4/3ms^-2#

Therefore,

#sqrt105=0+1/2*4/3*t^2#

#t^2=3/2*sqrt105#

#t=sqrt(3/2*sqrt105)=3.92s#

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Answer 2

Julia Aguilar

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,
( u ) is the initial velocity (which is zero since the object is at rest),
( a ) is the acceleration,
( t ) is the time.

Rearrange the equation to solve for time:

[ t = \sqrt{\frac{2s}{a}} ]

Plug in the values: [ t = \sqrt{\frac{2 \sqrt{(3-8)^2 + (1-9)^2 + (6-2)^2}}{\frac{4}{3}}} ]

Calculate the result.

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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.