An object is at rest at $(7 ,3 ,9 )$ and constantly accelerates at a rate of $7/4 m/s^2$ as it moves to point B. If point B is at $(2 ,5 ,0 )$ , 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 · 2018-06-10T06:15:33

The time is $=3.46s$

The distance $AB$ is

$||vec(AB)||=sqrt((2-7)^2+(5-3)^2+(0-9)^2)=sqrt(25+4+81)=sqrt(110)m$

Apply the equation of motion

$s=ut+1/2at^2$

The initial velocity is $u=0ms^-1$

The acceleration is $a=7/4ms^-2$

Therefore,

$sqrt110=0+1/2*7/4*t^2$

$t^2=8/7sqrt110$

$t=sqrt(8/7sqrt110)=3.46s$

The time is $=3.46s$

An object is at rest at (7 ,3 ,9 )(7 ,3 ,9 ) and constantly accelerates at a rate of 7/4 m/s^27/4 m/s^2 as it moves to point B. If point B is at (2 ,5 ,0 )(2 ,5 ,0 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.