An object is at rest at $(1, 5, 5)$ $(1 ,5 ,5 )$ and constantly accelerates at a rate of $\frac{7}{4}$ $7/4$ $m s^{- 2}$ $ms^-2$ as it moves to point B. If point B is at $(2, 3, 6)$ $(2 ,3 ,6 )$ , 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 $(1, 5, 5)$ $(1 ,5 ,5 )$ and constantly accelerates at a rate of $\frac{7}{4}$ $7/4$ $m s^{- 2}$ $ms^-2$ as it moves to point B. If point B is at $(2, 3, 6)$ $(2 ,3 ,6 )$ , 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 · 2016-08-06T08:51:52

First find the distance between the points, then use $s = u t + \frac{1}{2} a t^{2}$ $s=ut+1/2at^2$ to find the time taken.

Explanation:

First find the distance between the points:

$r = \sqrt{{(2 - 1)}^{2} + {(3 - 5)}^{2} + {(6 - 1)}^{2}} = \sqrt{6} \approx 2.45$ $r=sqrt((2-1)^2+(3-5)^2+(6-1)^2)=sqrt6~~2.45$ $m$ $m$

Now use the formula:

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

The initial velocity, $u = 0$ $u=0$ $m s^{- 1}$ $ms^-1$ , so $s = \frac{1}{2} a t^{2}$ $s=1/2at^2$ . We can rearrange to make $t$ $t$ the subject:

$t = \sqrt{\frac{2 s}{a}} = \sqrt{\frac{2 \times 2.45}{\frac{7}{4}}} = \sqrt{2.8} \approx 1.67$ $t=sqrt((2s)/a)=sqrt((2xx2.45)/(7/4))=sqrt(2.8)~~1.67$ $s$ $s$

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An object is at rest at $(1, 5, 5)$ $(1 ,5 ,5 )$ and constantly accelerates at a rate of $\frac{7}{4}$ $7/4$ $m s^{- 2}$ $ms^-2$ as it moves to point B. If point B is at $(2, 3, 6)$ $(2 ,3 ,6 )$ , 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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1 Answer

Explanation:

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An object is at rest at $(1, 5, 5)$ $(1 ,5 ,5 )$ and constantly accelerates at a rate of $\frac{7}{4}$ $7/4$ $m s^{- 2}$ $ms^-2$ as it moves to point B. If point B is at $(2, 3, 6)$ $(2 ,3 ,6 )$ , how long will it take for the object to reach point B? Assume that all coordinates are in meters.