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.

1 Answer
Nathan L.
May 19, 2017

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.

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