An object is at rest at (2 ,1 ,5 ) and constantly accelerates at a rate of 3 m/s as it moves to point B. If point B is at (6 ,7 ,5 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.

1 Answer
Shwetank Mauria
Jun 24, 2016

It will take 2.193 seconds.

Explanation:

The distance between two points (x_1,y_1,z_1) and (x_2,y_2,z_2) is given by

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

Hence distance between (2,1,5) and (6,7,5) is

sqrt((6-2)^2+(7-1)^2+(5-5)^2)

= sqrt(4^2+6^2+0^2)=sqrt(16+36+0)=sqrt52=2sqrt13

(As distance covered is given by S=ut+1/2at^2, where u is initial velocity, a is accelaration and t is time taken. If body is at rest S=1/2at^2 and hence t=sqrt((2S)/a)

As the coordinates are in meters, the time taken at an acceleration of 3 m/sec^2 will be given by

t=sqrt((2xx2sqrt13)/3)=sqrt((4xx3.606)/3)=sqrt(4.808)=2.193

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