An object is at rest at $(1, 2, 9)$ $(1 ,2 ,9 )$ and constantly accelerates at a rate of $1 \frac{m}{s^{2}}$ $1 m/s^2$ as it moves to point B. If point B is at $(3, 1, 4)$ $(3 ,1 ,4 )$ , 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 $(1, 2, 9)$ $(1 ,2 ,9 )$ and constantly accelerates at a rate of $1 \frac{m}{s^{2}}$ $1 m/s^2$ as it moves to point B. If point B is at $(3, 1, 4)$ $(3 ,1 ,4 )$ , 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 · 2017-05-23T01:06:44

$3.31 s$ $3.31"s"$

Explanation:

We're dealing with one-dimensional motion with constant acceleration.

What we can do first is find the displacement $s$ $s$ during the time interval from point A to point B by using the three-dimensional distance formula:

$s = \sqrt{{(3 m - 1 m)}^{2} + {(1 m - 2 m)}^{2} + {(4 m - 9 m)}^{2}} = 5.48 m$ $s = sqrt((3"m"-1"m")^2 + (1"m"-2"m")^2 + (4"m"-9"m")^2) = 5.48"m"$

Let's look at our known quantities:

Since the object is initially at rest, $v_{0 x} = 0$ $v_(0x) = 0$

We'll make the initial position $x_{0}$ $x_0$ be $0$ $0$ , and the position $x$ $x$ at time $t$ $t$ be $5.48 m$ $5.48"m"$

The acceleration is constant at $1 \frac{m}{s}$ $1"m"/"s"$

To find the time duration $t$ $t$ of the displacement $s$ $s$ , we can use the equation

$x = x_{0} + v_{0 x} t + \frac{1}{2} a_{x} t^{2}$ $x = x_0 + v_(0x)t + 1/2a_xt^2$

And since the initial velocity and position are both $0$ $0$ , the equation becomes

$x = \frac{1}{2} a_{x} t^{2}$ $x = 1/2a_xt^2$

Here, the variable $x$ $x$ is analogous to the displacement variable $s$ $s$ , since they both represent the distance covered. Plugging in the known variables $a_{x}$ $a_x$ and $x$ $x$ , the time $t$ $t$ elapsed during the displacement is

$5.48 m = \frac{1}{2} (1 \frac{m}{s^{2}}) t^{2}$ $5.48"m" = 1/2(1"m"/("s"^2))t^2$

$t = \sqrt{\frac{2 (5.48 \frac{m}{s})}{1 \frac{m}{s^{2}}}} = 3.31 s$ $t = sqrt((2(5.48"m"/"s"))/(1"m"/("s"^2))) = color(red)(3.31"s"$

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An object is at rest at $(1, 2, 9)$ $(1 ,2 ,9 )$ and constantly accelerates at a rate of $1 \frac{m}{s^{2}}$ $1 m/s^2$ as it moves to point B. If point B is at $(3, 1, 4)$ $(3 ,1 ,4 )$ , how long will it take for the object to reach point B? Assume that all coordinates are in meters.

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Explanation:

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An object is at rest at (1,2,9)(1 ,2 ,9 ) and constantly accelerates at a rate of 1ms21 m/s^2 as it moves to point B. If point B is at (3,1,4)(3 ,1 ,4 ), 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 $(1, 2, 9)$ $(1 ,2 ,9 )$ and constantly accelerates at a rate of $1 \frac{m}{s^{2}}$ $1 m/s^2$ as it moves to point B. If point B is at $(3, 1, 4)$ $(3 ,1 ,4 )$ , how long will it take for the object to reach point B? Assume that all coordinates are in meters.