Question

An object is at rest at `(7 ,6 ,4 )` and constantly accelerates at a rate of `5/4 m/s^2` as it moves to point B. If point B is at `(9 ,5 ,7 )`, how long will it take for the object to reach point B? Assume that all coordinates are in meters.

Answer 1

It will take `2.45` 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 `(7,6,4)` and `(9,5,7)` is

`sqrt((9-7)^2+(5-6)^2+(7-4)^2)`

= `sqrt(2^2+1^2+3^2)=sqrt(4+1+9)=sqrt14`

(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 `5/4` `m/sec^2` will be given by

`t=sqrt((2sqrt14)/(5/4))=sqrt((2sqrt14)xx(4/5))=sqrt((8sqrt14)/5))`

= `sqrt(8xx3.74166/5)=sqrt5.986656=2.45`