An object is at rest at (8 ,3 ,3 ) and constantly accelerates at a rate of 7/4 ms^-2 as it moves to point B. If point B is at (6 ,3 ,4 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.

1 Answer
David G.
Jul 8, 2017

First step is to find the distance between the two points: r~=2.24 m

Second step is to find the time taken: t=1.6 s

Explanation:

To find the distance between two points in 3D space we have:

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

=sqrt((6-8)^2+(3-3)^2+(4-3)^2)=sqrt((-2)^2+(0)^2+(1)^2)

Therefore:

r=sqrt5~=2.24

Now to find the time taken, we can use:

d=ut+1/2at^2

Since the object is initially at rest, the ut term goes to 0.

Rearranging:

t=sqrt((2d)/a)

Note that I have used both d and r because of the way I remember formulae, but we're talking about the same distance - the distance between the points.

t=sqrt((2xx2.24)/(7/4))= 1.6 s

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