An object is at rest at $(7, 6, 4)$ $(7 ,6 ,4 )$ 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 $(9, 5, 7)$ $(9 ,5 ,7 )$ , how long will it take for the object to reach point B? Assume that all coordinates are in meters.

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An object is at rest at $(7, 6, 4)$ $(7 ,6 ,4 )$ 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 $(9, 5, 7)$ $(9 ,5 ,7 )$ , 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 · 2016-06-23T14:31:05

It will take $2.45$ $2.45$ seconds.

Explanation:

The distance between two points $(x_{1}, y_{1}, z_{1})$ $(x_1,y_1,z_1)$ and $(x_{2}, y_{2}, z_{2})$ $(x_2,y_2,z_2)$ is given by

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

Hence distance between $(7, 6, 4)$ $(7,6,4)$ and $(9, 5, 7)$ $(9,5,7)$ is

$\sqrt{{(9 - 7)}^{2} + {(5 - 6)}^{2} + {(7 - 4)}^{2}}$ $sqrt((9-7)^2+(5-6)^2+(7-4)^2)$

= $\sqrt{2^{2} + 1^{2} + 3^{2}} = \sqrt{4 + 1 + 9} = \sqrt{14}$ $sqrt(2^2+1^2+3^2)=sqrt(4+1+9)=sqrt14$

(As distance covered is given by $S = u t + \frac{1}{2} a t^{2}$ $S=ut+1/2at^2$ , where $u$ $u$ is initial velocity, $a$ $a$ is accelaration and $t$ $t$ is time taken. If body is at rest $S = \frac{1}{2} a t^{2}$ $S=1/2at^2$ and hence $t = \sqrt{\frac{2 S}{a}}$ $t=sqrt((2S)/a)$

As the coordinates are in meters, the time taken at an acceleration of $\frac{5}{4}$ $5/4$ $\frac{m}{{sec}^{2}}$ $m/sec^2$ will be given by

$t = \sqrt{\frac{2 \sqrt{14}}{\frac{5}{4}}} = \sqrt{(2 \sqrt{14}) \times (\frac{4}{5})} = \sqrt{\frac{8 \sqrt{14}}{5}} ⎞ ⎠$ $t=sqrt((2sqrt14)/(5/4))=sqrt((2sqrt14)xx(4/5))=sqrt((8sqrt14)/5))$

= $\sqrt{8 \times \frac{3.74166}{5}} = \sqrt{5.986656} = 2.45$ $sqrt(8xx3.74166/5)=sqrt5.986656=2.45$

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An object is at rest at $(7, 6, 4)$ $(7 ,6 ,4 )$ 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 $(9, 5, 7)$ $(9 ,5 ,7 )$ , how long will it take for the object to reach point B? Assume that all coordinates are in meters.

1 Answer

Explanation:

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An object is at rest at (7,6,4)(7 ,6 ,4 ) and constantly accelerates at a rate of 54ms25/4 m/s^2 as it moves to point B. If point B is at (9,5,7)(9 ,5 ,7 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.

1 Answer

Explanation:

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An object is at rest at $(7, 6, 4)$ $(7 ,6 ,4 )$ 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 $(9, 5, 7)$ $(9 ,5 ,7 )$ , how long will it take for the object to reach point B? Assume that all coordinates are in meters.