An object is at rest at $(3, 8, 8)$ $(3 ,8 ,8 )$ and constantly accelerates at a rate of $\frac{7}{5} \frac{m}{s^{2}}$ $7/5 m/s^2$ as it moves to point B. If point B is at $(2, 9, 1)$ $(2 ,9 ,1 )$ , 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 · 2017-06-18T17:13:54

The time is $= 3.2 s$ $=3.2s$

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

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

$d = A B = \sqrt{{(2 - 3)}^{2} + {(9 - 8)}^{2} + {(1 - 8)}^{2}}$ $d=AB= sqrt((2-3)^2+(9-8)^2+(1-8)^2)$

$= \sqrt{1^{2} + 1^{2} + 7^{2}}$ $=sqrt(1^2+1^2+7^2)$

$= \sqrt{1 + 1 + 49}$ $=sqrt(1+1+49)$

$= \sqrt{51}$ $=sqrt51$

$= 7.14 m$ $=7.14m$

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,

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

$a = \frac{7}{5} m s^{- 2}$ $a=7/5ms^-2$

$t^{2} = \frac{2 d}{a} = \frac{2 \cdot 7.14}{\frac{7}{5}}$ $t^2=(2d)/a=(2*7.14)/(7/5)$

$= 10.2 s^{2}$ $=10.2s^2$

$t = \sqrt{10.2} = 3.2 s$ $t=sqrt(10.2)=3.2s$

An object is at rest at (3,8,8)(3 ,8 ,8 ) and constantly accelerates at a rate of 75ms27/5 m/s^2 as it moves to point B. If point B is at (2,9,1)(2 ,9 ,1 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.