An object is at rest at (6 ,5 ,9 ) and constantly accelerates at a rate of 4/3 m/s^2 as it moves to point B. If point B is at (3 ,6 ,4 ), 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 2.979 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 (6,5,9) and (3,6,4) is

sqrt((3-6)^2+(6-5)^2+(4-9)^2)

= sqrt(3^2+1^2+5^2)=sqrt(9+1+25)=sqrt35=5.916

(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 4/3 m/sec^2 will be given by

t=sqrt((2xx5.916)/(4/3))=sqrt((6xx5.916)/4)=sqrt(8.874)=2.979

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