Question

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.

Answer 1

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`