An object is at rest at (8 ,6 ,2 ) and constantly accelerates at a rate of 1/4 m/s^2 as it moves to point B. If point B is at (2 ,8 ,2 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.

1 Answer
Shwetank Mauria
Jul 1, 2016

It will take 7.11 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 (8,6,2) and (2,8,2) is

sqrt((2-8)^2+(8-6)^2+(2-2)^2)

= sqrt(6^2+2^2+0^2)=sqrt(36+4+0)=sqrt40

(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 1/4 m/sec^2 will be given by

t=sqrt((2xxsqrt40)/(1/4))=sqrt(4xx2xx6.3246)=sqrt(17.888)=7.11

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