An object is at rest at $(2, 1, 3)$ $(2 ,1 ,3 )$ 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 $(9, 3, 5)$ $(9 ,3 ,5 )$ , 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 · 2017-03-27T18:41:56

The time is $= 3.89 s$ $=3.89s$

The distance between the 2 points $A = (2, 1, 3)$ $A=(2,1,3)$ and $B = (9, 3, 5)$ $B=(9,3,5)$ is

$d = \sqrt{{(9 - 2)}^{2} + {(3 - 1)}^{2} + {(5 - 3)}^{2}}$ $d=sqrt((9-2)^2+(3-1)^2+(5-3)^2)$

$= \sqrt{49 + 4 + 4}$ $=sqrt(49+4+4)$

$= \sqrt{57}$ $=sqrt57$

$a (t) = 1$ $a(t)=1$

The velocity is

$v (t) = \int_{0}^{t} 1 d t = t$ $v(t)=int_0^t1dt=t$

The distance is

$s = \int_{0}^{t} t d t = \frac{1}{2} t^{2}$ $s=int_0^t tdt=1/2t^2$

This is equal to

sqrt57=1/2t^2

$t^{2} = 2 \sqrt{57} = 15.1$ $t^2=2sqrt57=15.1$

So,

$t = \sqrt{15.1} = 3.89 s$ $t=sqrt15.1=3.89s$

An object is at rest at (2,1,3)(2 ,1 ,3 ) and constantly accelerates at a rate of 1ms21 m/s^2 as it moves to point B. If point B is at (9,3,5)(9 ,3 ,5 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.