An object is at rest at $(7, 8, 4)$ $(7 ,8 ,4 )$ 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 $(1, 5, 3)$ $(1 ,5 ,3 )$ , how long will it take for the object to reach point B? Assume that all coordinates are in meters.

Question

An object is at rest at $(7, 8, 4)$ $(7 ,8 ,4 )$ 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 $(1, 5, 3)$ $(1 ,5 ,3 )$ , how long will it take for the object to reach point B? Assume that all coordinates are in meters.

1 Answer

Impact of this question

1522 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-05-28T09:38:11

It takes 3.29 seconds.

Explanation:

First of all we need to know how much distance the objects has to travel. The distance between two points is given by the Pitagora's theorem in 3 dimensions

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

in our case

$d = \sqrt{{(7 - 1)}^{2} + {(8 - 5)}^{2} + {(4 - 3)}^{2}} = \sqrt{36 + 9 + 1} = \sqrt{46} \approx 6.78 m$ $d=sqrt((7-1)^2+(8-5)^2+(4-3)^2)=sqrt(36+9+1)=sqrt(46)\approx6.78m$

Now that we know that our body will travel for 6.78 m we can use the equation of motion for an accelerating body that is

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

where $x$ $x$ is the traveled distance, for us 6.78, $a$ $a$ is the acceleration, for us $\frac{5}{4} \frac{m}{s^{2}}$ $5/4m/s^2$ and $t$ $t$ is the time that we want to find. I plug the values in the equation

$6.78 = \frac{1}{2} \cdot \frac{5}{4} \cdot t^{2}$ $6.78=1/2*5/4*t^2$
$6.78 = \frac{5}{8} t^{2}$ $6.78=5/8t^2$

I am interested in $t$ $t$ so I multiply both sides for $\frac{8}{5}$ $8/5$ and do the square root

$\sqrt{6.78 \cdot \frac{8}{5}} = t$ $sqrt(6.78*8/5)=t$
$t = \sqrt{10.848} \approx 3.29$ $t=sqrt(10.848)\approx 3.29$ s.

Science

Math

Humanities

... and beyond

An object is at rest at $(7, 8, 4)$ $(7 ,8 ,4 )$ 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 $(1, 5, 3)$ $(1 ,5 ,3 )$ , how long will it take for the object to reach point B? Assume that all coordinates are in meters.

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

An object is at rest at (7,8,4)(7 ,8 ,4 ) and constantly accelerates at a rate of 54ms25/4 m/s^2 as it moves to point B. If point B is at (1,5,3)(1 ,5 ,3 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.

1 Answer

Explanation:

Related questions

Impact of this question

An object is at rest at $(7, 8, 4)$ $(7 ,8 ,4 )$ 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 $(1, 5, 3)$ $(1 ,5 ,3 )$ , how long will it take for the object to reach point B? Assume that all coordinates are in meters.