An object is at rest at $(8, 1, 2)$ $(8 ,1 ,2 )$ and constantly accelerates at a rate of $3 \frac{m}{s}$ $3 m/s$ as it moves to point B. If point B is at $(6, 7, 5)$ $(6 ,7 ,5 )$ , 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 · 2017-04-26T06:43:33

The time is $= 2.16 s$ $=2.16s$

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

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

$d = A B = \sqrt{{(6 - 8)}^{2} + {(7 - 1)}^{2} + {(5 - 2)}^{2}}$ $d=AB= sqrt((6-8)^2+(7-1)^2+(5-2)^2)$

$= \sqrt{2^{2} + 6^{2} + 3^{2}}$ $=sqrt(2^2+6^2+3^2)$

$= \sqrt{4 + 36 + 9}$ $=sqrt(4+36+9)$

$= \sqrt{49}$ $=sqrt49$

$= 7$ $=7$

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,

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

$t^{2} = \frac{2 d}{a} = \frac{2 \cdot 7}{3}$ $t^2=(2d)/a=(2*7)/(3)$

$= 4.67$ $=4.67$

$t = \sqrt{4.67} = 2.16 s$ $t=sqrt(4.67)=2.16s$

An object is at rest at (8,1,2)(8 ,1 ,2 ) and constantly accelerates at a rate of 3ms3 m/s as it moves to point B. If point B is at (6,7,5)(6 ,7 ,5 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.