An object is at rest at $(1, 7, 4)$ $(1 ,7 ,4 )$ and constantly accelerates at a rate of $\frac{5}{4} \frac{m}{s^{2}}$ $5/4 m/s^2$ as it moves to point B. If point B is at $(8, 5, 6)$ $(8 ,5 ,6 )$ , 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 · 2017-06-07T06:24:47

The time is $= 3.48 s$ $=3.48s$

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

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

$d = A B = \sqrt{{(8 - 1)}^{2} + {(5 - 7)}^{2} + {(6 - 4)}^{2}}$ $d=AB= sqrt((8-1)^2+(5-7)^2+(6-4)^2)$

$= \sqrt{7^{2} + 2^{2} + 2^{2}}$ $=sqrt(7^2+2^2+2^2)$

$= \sqrt{49 + 4 + 4}$ $=sqrt(49+4+4)$

$= \sqrt{57}$ $=sqrt57$

$= 7.55 m$ $=7.55m$

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,

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

$a = \frac{5}{4} m s^{- 2}$ $a=5/4ms^-2$

$t^{2} = \frac{2 d}{a} = \frac{2 \cdot 7.55}{\frac{5}{4}}$ $t^2=(2d)/a=(2*7.55)/(5/4)$

$= 12.08 s^{2}$ $=12.08s^2$

$t = \sqrt{12.08} = 3.48 s$ $t=sqrt(12.08)=3.48s$

An object is at rest at (1,7,4)(1 ,7 ,4 ) and constantly accelerates at a rate of 54ms25/4 m/s^2 as it moves to point B. If point B is at (8,5,6)(8 ,5 ,6 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.