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

Answer 1

Emily Abbate

The time is #=2.65s#

The distance #AB# is

#AB=sqrt((5-7)^2+(1-6)^2+(7-4)^2)=sqrt(4+25+9)=sqrt(38)m#

Apply the equation of motion

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

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

The acceleration is #a=7/4ms^-2#

Therefore,

#sqrt38=0+1/2*7/4*t^2#

#t^2=8/7*sqrt38#

#t=sqrt(8/7*sqrt38)=2.65s#

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Answer 2

Leo Abeyta

To calculate the time it takes for the object to reach point B, use the kinematic equation:

[d = ut + \frac{1}{2}at^2]

Where:

(d) is the displacement (distance between points A and B)
(u) is the initial velocity (assumed to be zero as the object is at rest)
(a) is the acceleration
(t) is the time

Rearrange the equation to solve for (t):

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

Substitute the values:

[d = \sqrt{(5-7)^2 + (1-6)^2 + (7-4)^2}] [a = \frac{7}{4}]

Calculate (t).

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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.