Question

An object is at rest at `(1 ,7 ,2 )` and constantly accelerates at a rate of `1 m/s^2` as it moves to point B. If point B is at `(3 ,1 ,4 )`, how long will it take for the object to reach point B? Assume that all coordinates are in meters.

Answer 1

`2.12s`

Explanation:

We can solve this just a like a one-dimensional constant acceleration problem; the distance covered is

`Deltax = sqrt (3^2 + 1^2 + 4^2) - sqrt(1^2 + 7^2 + 2^2)= -2.249 m`

And since the acceleration `a_x = 1m/(s^2)`, the time `t` it takes the object to travel `2.249m` (in the reverse direction) is

`t = sqrt((2Deltax)/(a_x)) = sqrt((2(2.249m))/(1m/(s^2))) = 2.12 s`

This is derived from the equation `x = x_0 + v_(0x)t + 1/2a_xt^2`, where `v_(0x)` is `0` (initially at rest), and I made `x` the distance `2.249m` (sign is arbitrary, but needs to be positive to yield a real answer) and `x_0` `0`.