An object is at rest at #(3 ,0 ,3 )# and constantly accelerates at a rate of #7/5 m/s^2# as it moves to point B. If point B is at #(5 ,3 ,5 )#, how long will it take for the object to reach point B? Assume that all coordinates are in meters.

Answer 1

William Audy

The time is #=2.43s#

The distance #AB# is

#s=sqrt((5-3)^2+(3-0)^2+(5-3)^2)#

#=sqrt(4+9+4)#

#=sqrt17m#

We apply the equation of motion

#v^2=u^2+2as#

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

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

Therefore,

#v^2=0+2*7/5*sqrt17#

#v^2=11.54#

#v=sqrt(11.54)=3.4ms^-1#

To determine the time, we apply the equation of motion

#v=u+at#

#t=v/a#

#t=3.4/(7/5)=2.43s#

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Answer 2

Xavier Adame

It will take approximately 2.24 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.