An object is at rest at $(1, 7, 2)$ $(1 ,7 ,2 )$ 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 $(3, 7, 4)$ $(3 ,7 ,4 )$ , how long will it take for the object to reach point B? Assume that all coordinates are in meters.

Question

1611 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-22T06:34:00

The answer is $= 2.4 s$ $=2.4s$

The distance between 2 points A $(x_{A}, y_{A}, z_{A})$ $(x_A,y_A,z_A)$ and

B $(x_{B}, y_{B}, z_{B})$ $(x_B,y_B, z_B)$ is

$d = \sqrt{{(x_{B} - x_{A})}_{B}^{2} + {(y_{B} - y_{A})}_{B}^{2} + {(z_{B} - z_{A})}_{B}^{2}}$ $d=sqrt((x_B-x_A)^2+(y_B-y_A)^2+(z_B-z_A)^2)$

Here we have, A $(1, 7, 2)$ $(1,7,2)$ and B $(3, 7, 4)$ $(3,7,4)$

so,

$d = \sqrt{{(3 - 1)}^{2} + {(7 - 7)}^{2} + {(4 - 2)}^{2}}$ $d=sqrt((3-1)^2+(7-7)^2+(4-2)^2)$

$= \sqrt{4 + 4} = \sqrt{8} m$ $=sqrt(4+4)=sqrt8m$

We use the equation

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

As, the object is initially at rest $u = 0$ $u=0$

$a = 1 m s^{- 2}$ $a=1ms^(-2)$

Then,

$\sqrt{8} = \frac{1}{2} \cdot 1 \cdot t^{2}$ $sqrt8=1/2*1*t^2$

$t^{2} = 2 \sqrt{8}$ $t^2=2sqrt8$

$t = \sqrt{2 \sqrt{8}} = 2.4 s$ $t=sqrt(2sqrt8)=2.4s$

An object is at rest at (1,7,2)(1 ,7 ,2 ) and constantly accelerates at a rate of 1ms21 m/s^2 as it moves to point B. If point B is at (3,7,4)(3 ,7 ,4 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.