An object is at rest at $(7, 9, 4)$ $(7 ,9 ,4 )$ 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 $(7, 1, 6)$ $(7 ,1 ,6 )$ , 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 $(7, 9, 4)$ $(7 ,9 ,4 )$ 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 $(7, 1, 6)$ $(7 ,1 ,6 )$ , 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 · 2018-03-04T12:34:44

$3.071 s$ $3.071s$

Explanation:

In this pathway,net displacement of the object is $\sqrt{{(7 - 1)}^{2} + {(1 - 9)}^{2} + {(6 - 4)}^{2}} = 8.25 m$ $sqrt((7-1)^2 +(1-9)^2 +(6-4)^2)=8.25m$

So,to calculate time required for this motion under constant acceleration $a$ $a$ we can use,

$s = u t + \frac{1}{2} a t^{2}$ $s=ut + 1/2 a t^2$

Here, $u = 0$ $u=0$ (as the object was initially at rest)

and, $a = \frac{7}{4}, s = 8.25$ $a=7/4,s=8.25$

So, $t = 3.071 s$ $t=3.071s$

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An object is at rest at $(7, 9, 4)$ $(7 ,9 ,4 )$ 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 $(7, 1, 6)$ $(7 ,1 ,6 )$ , 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 (7,9,4)(7 ,9 ,4 ) and constantly accelerates at a rate of 74ms27/4 m/s^2 as it moves to point B. If point B is at (7,1,6)(7 ,1 ,6 ), 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 $(7, 9, 4)$ $(7 ,9 ,4 )$ 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 $(7, 1, 6)$ $(7 ,1 ,6 )$ , how long will it take for the object to reach point B? Assume that all coordinates are in meters.