An object is at rest at $(1, 5, 5)$ $(1 ,5 ,5 )$ and constantly accelerates at a rate of $\frac{4}{3} \frac{m}{s^{2}}$ $4/3 m/s^2$ as it moves to point B. If point B is at $(3, 6, 4)$ $(3 ,6 ,4 )$ , how long will it take for the object to reach point B? Assume that all coordinates are in meters.

Question

1345 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-11-26T07:34:26

The time is $= 1.92 s$ $=1.92s$

The distance $A B$ $AB$ is

$A B = \sqrt{{(3 - 1)}^{2} + {(6 - 5)}^{2} + {(4 - 5)}^{2}} = \sqrt{4 + 1 + 1} = \sqrt{6} m$ $AB=sqrt((3-1)^2+(6-5)^2+(4-5)^2)=sqrt(4+1+1)=sqrt6m$

Apply the equation of motion

$s = u t + \frac{1}{2} a t^{2}$ $s=ut+1/2at^2$

The initial velocity is $u = 0 m s^{- 1}$ $u=0ms^-1$

The acceleration is $a = \frac{4}{3} m s^{- 2}$ $a=4/3ms^-2$

Therefore,

$\sqrt{6} = 0 \cdot \frac{1}{2} \cdot \frac{4}{3} \cdot t^{2}$ $sqrt6=0*1/2*4/3*t^2$

$t^{2} = \frac{3}{2} \sqrt{6}$ $t^2=3/2sqrt6$

$t = \sqrt{\frac{3}{2} \sqrt{6}} = 1.92 s$ $t=sqrt(3/2sqrt6)=1.92s$

An object is at rest at (1,5,5)(1 ,5 ,5 ) and constantly accelerates at a rate of 43ms24/3 m/s^2 as it moves to point B. If point B is at (3,6,4)(3 ,6 ,4 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.