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.

Question

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.

Emily Abbate · Accepted Answer

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#

Layla Ahmed · Answer

undefined To calculate the time it takes for the object to reach point B, you can use the kinematic equation for uniformly accelerated motion in three dimensions:

$$ \vec{r} = \vec{r}_0 + \vec{v}_0 t + \frac{1}{2} \vec{a} t^2 $$

Where:
- $ \vec{r} $ is the final position vector (point B).
- $ \vec{r}_0 $ is the initial position vector (point A).
- $ \vec{v}_0 $ is the initial velocity vector (initially at rest, so $ \vec{v}_0 = 0 $).
- $ \vec{a} $ is the acceleration vector.
- $ t $ is the time taken.

Given:
- $ \vec{r}_0 = (7, 3, 9) $
- $ \vec{r} = (2, 5, 0) $
- $ \vec{a} = \left(0, 0, \frac{7}{4}\right) $

You can solve for $ t $ by substituting the known values into the equation and solving for $ t $.