An object is at rest at #(5 ,2 ,8 )# and constantly accelerates at a rate of #2/5# #ms^-2# as it moves to point B. If point B is at #(6 ,3 ,2 )#, 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 #(5  ,2   ,8    )# and constantly accelerates at a rate of #2/5#  #ms^-2# as it moves to point B.  If point B is at #(6 ,3 ,2   )#, how long will it take for the object to reach point B? Assume that all coordinates are in meters.

Nora Arzate · Accepted Answer

First we find the distance between the two points:

#d=sqrt((x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2)#
#d=sqrt((6-5)^2+(3-2)^2+(2-8)^2)#
#d=sqrt(1+1+36)=sqrt(38)#

Then we know #d=ut+1/2at^2#. We know #u=0#, and rearranging we get:

#t=sqrt((2d)/a)=sqrt((2xxsqrt(38))/(2/5))~=5.55# #s#

Elena Albarran · Answer

The time is #=5.55s#

The distance #AB# is

#AB=sqrt((6-5)^2+(3-2)^2+(2-8)^2)=sqrt(1+1+36)=sqrt(38)m#

Apply the equation of motion

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

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

The acceleration is #a=2/5ms^-2#

Therefore,

#sqrt38=0+1/2*2/5*t^2#

#t^2=5*sqrt38#

#t=sqrt(5*sqrt38)=5.55s#

William Audy · Answer

undefined To find the time it takes for the object to reach point B, you can use the kinematic equation:

$$ \Delta x = v_i t + \frac{1}{2} a t^2 $$

Where:
- $ \Delta x $ is the displacement between the initial and final positions.
- $ v_i $ is the initial velocity (which is 0 since the object is at rest).
- $ a $ is the acceleration.
- $ t $ is the time.

Given:
- Initial position: $ (5, 2, 8) $
- Final position: $ (6, 3, 2) $
- Acceleration: $ 2/5 \, \text{m/s}^2 $

Substituting the values:

$$ \sqrt{(6-5)^2 + (3-2)^2 + (2-8)^2} = \frac{1}{2} \left( \frac{2}{5} \right) t^2 $$

$$ \sqrt{1^2 + 1^2 + (-6)^2} = \frac{1}{2} \left( \frac{2}{5} \right) t^2 $$

$$ \sqrt{1 + 1 + 36} = \frac{1}{2} \left( \frac{2}{5} \right) t^2 $$

$$ \sqrt{38} = \frac{1}{2} \left( \frac{2}{5} \right) t^2 $$

$$ t^2 = \frac{2 \cdot \sqrt{38}}{5} $$

$$ t = \sqrt{\frac{2 \cdot \sqrt{38}}{5}} $$

$$ t \approx 1.95 \, \text{s} $$

So, it will take approximately 1.95 seconds for the object to reach point B.