An object is at rest at (2 ,1 ,6 ) and constantly accelerates at a rate of 1/4 m/s^2 as it moves to point B. If point B is at (3 ,2 ,7 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.

1 Answer
David G.
May 21, 2017

t = sqrt((2s)/a) = sqrt ((2xx1.73)/0.25) = 3.72 s

Explanation:

In a 3-dimensional coordinate system, the distance between two points is given by:

r=sqrt((x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2)

r=sqrt((3-2)^2+(2-1)^2+(7-6)^2) =sqrt3 = 1.73 m

We can use the equation of motion:

s = ut + 1/2 at^2

Since the initial velocity u=0, we can ignore the first term, and rearrange s= 1/2 at^2 to make t the subject.

Note that the term 'r' I was using in the distance calculation above is the same as the distance 's' that the object moves. I don't want the symbol change to be confusing.

t = sqrt((2s)/a) = sqrt ((2xx1.73)/0.25)

(I used 0.25 rather than 1/4 for ease of writing, but of course it means the same thing)

Therefore t = 3.72 s.

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