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

Question

1374 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-03-30T12:47:14

$t = 3, 43 s$ $t=3,43 "s"$

$distance between (1,7,4) and (8,5,3):$ $"distance between (1,7,4) and (8,5,3):"$

$s = \sqrt{{(8 - 1)}^{2} + {(5 - 7)}^{2} + {(3 - 4)}^{2}}$ $s=sqrt((8-1)^2+(5-7)^2+(3-4)^2)$

$s = \sqrt{7^{2} + {(- 2)}^{2} + {(- 1)}^{2}}$ $s=sqrt(7^2+(-2)^2+(-1)^2)$

$s = \sqrt{49 + 4 + 1}$ $s=sqrt(49+4+1)$

$s = \sqrt{54}$ $s=sqrt54$

$s = 7, 35 m$ $s=7,35 " m"$

$a = \frac{5}{4} = 1, 25 \frac{m}{s^{2}}$ $a=5/4=1,25 " "m/s^2$

$t = ?$ $t=?$

$s = \frac{1}{2} \cdot a \cdot t^{2}$ $s=1/2*a*t^2$

$t^{2} = 2 \cdot \frac{s}{a}$ $t^2=2*s/a$

$t^{2} = \frac{2 \cdot 7, 35}{1, 25}$ $t^2=(2*7,35)/(1,25)$

$t^{2} = 11, 76$ $t^2=11,76$

$\sqrt{t^{2}} = \sqrt{11, 76}$ $sqrt (t^2)=sqrt (11,76)$

$t = 3, 43 s$ $t=3,43 "s"$

An object is at rest at (1 ,7 ,4 )(1,7,4)(1 ,7 ,4 ) and constantly accelerates at a rate of 5/4 m/s^254ms25/4 m/s^2 as it moves to point B. If point B is at (8 ,5 ,3 )(8,5,3)(8 ,5 ,3 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.