An object is at rest at $(8, 2, 5)$ $(8 ,2 ,5 )$ and constantly accelerates at a rate of $\frac{7}{4} \frac{m}{s^{2}}$ $7/4 m/s^2$ as it moves to point B. If point B is at $(2, 3, 7)$ $(2 ,3 ,7 )$ , 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 · 2017-03-18T15:03:20

The time taken is $= 2.71 s$ $=2.71s$

We are going to apply the following equation of motion :

$s = u_{0} t + \frac{1}{2} a t^{2}$ $s=u_0t+1/2at^2$

$u_{0} = 0$ $u_0=0$

So,

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

$a = \frac{7}{4} m s^{- 2}$ $a=7/4ms^-2$

The distance between 2 points $A = (x_{1}, y_{1}, z_{1})$ $A=(x_1,y_1,z_1)$ and $B = (x_{2} . y_{2} . z_{2})$ $B=(x_2.y_2.z_2)$ is

$= \sqrt{{(x_{2} - x_{1})}_{2}^{2} + {(y_{2} - y_{1})}_{2}^{2} + {(z_{2} - z_{1})}_{2}^{2}}$ $=sqrt((x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2)$

Here,

$A = (8, 2, 5)$ $A=(8,2,5)$ and $B = (2, 3, 7)$ $B=(2,3,7)$

So,

$s = \sqrt{{(2 - 8)}^{2} + {(3 - 2)}^{2} + {(7 - 5)}^{2}}$ $s=sqrt((2-8)^2+(3-2)^2+(7-5)^2)$

$= \sqrt{36 + 1 + 4}$ $=sqrt(36+1+4)$

$= \sqrt{41}$ $=sqrt41$

From the equation of motion,

$t^{2} = \frac{2 s}{a}$ $t^2=(2s)/a$

$t^{2} = \frac{2 \cdot \sqrt{41}}{\frac{7}{4}} = 7.32$ $t^2=(2*sqrt41)/(7/4)=7.32$

$t = \sqrt{7.32} = 2.71 s$ $t=sqrt7.32=2.71s$

An object is at rest at (8,2,5)(8 ,2 ,5 ) and constantly accelerates at a rate of 74ms27/4 m/s^2 as it moves to point B. If point B is at (2,3,7)(2 ,3 ,7 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.