An object is at rest at $(2, 7, 1)$ $(2 ,7 ,1 )$ 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 $(3, 1, 6)$ $(3 ,1 ,6 )$ , how long will it take for the object to reach point B? Assume that all coordinates are in meters.

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Answer 1 · 2017-05-26T08:25:36

The time is $= 3.44 s$ $=3.44s$

The distance between the points $A = (x_{A}, y_{A}, z_{A})$ $A=(x_A,y_A,z_A)$ and the point $B = (x_{B}, y_{B}, z_{B})$ $B=(x_B,y_B,z_B)$ is

$A B = \sqrt{{(x_{B} - x_{A})}_{B}^{2} + {(y_{B} - y_{A})}_{B}^{2} + {(z_{B} - z_{A})}_{B}^{2}}$ $AB=sqrt((x_B-x_A)^2+(y_B-y_A)^2+(z_B-z_A)^2)$

$d = A B = \sqrt{{(3 - 2)}^{2} + {(1 - 7)}^{2} + {(6 - 1)}^{2}}$ $d=AB= sqrt((3-2)^2+(1-7)^2+(6-1)^2)$

$= \sqrt{1^{2} + 6^{2} + 5^{2}}$ $=sqrt(1^2+6^2+5^2)$

$= \sqrt{1 + 36 + 25}$ $=sqrt(1+36+25)$

$= \sqrt{62}$ $=sqrt62$

$= 7.874 m$ $=7.874m$

We apply the equation of motion

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

$u = 0$ $u=0$

so,

$d = \frac{1}{2} a t^{2}$ $d=1/2at^2$

$t^{2} = \frac{2 d}{a} = \frac{2 \cdot 7.874}{\frac{4}{3}}$ $t^2=(2d)/a=(2*7.874)/(4/3)$

$= 11.81$ $=11.81$

$t = \sqrt{11.81} = 3.44 s$ $t=sqrt(11.81)=3.44s$

An object is at rest at (2,7,1)(2 ,7 ,1 ) and constantly accelerates at a rate of 43ms24/3 m/s^2 as it moves to point B. If point B is at (3,1,6)(3 ,1 ,6 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.