An object is at rest at #(8 ,3 ,8 )# and constantly accelerates at a rate of #7/4 m/s^2# as it moves to point B. If point B is at #(6 ,5 ,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 #(8 ,3   ,8    )# and constantly accelerates at a rate of #7/4  m/s^2# as it moves to point B.  If point B is at #(6  ,5   ,2  )#, how long will it take for the object to reach point B? Assume that all coordinates are in meters.

Amelia Adams · Accepted Answer

The time is #=2.75s#

The distance #AB# is

#AB=sqrt((6-8)^2+(5-3)^2+(2-8)^2)=sqrt(4+4+36)=(sqrt44)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,

#sqrt44=0+1/2*7/4*t^2#

#t^2=8/7sqrt44#

#t=sqrt(8/7sqrt44)=2.75s#

Charles Aeschliman · Answer

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

$$ \text{Distance} = \text{Initial velocity} \times \text{Time} + \frac{1}{2} \times \text{Acceleration} \times \text{Time}^2 $$

Given that the initial velocity is zero (object is at rest), the equation simplifies to:

$$ \text{Distance} = \frac{1}{2} \times \text{Acceleration} \times \text{Time}^2 $$

Rearrange the equation to solve for time:

$$ \text{Time} = \sqrt{\frac{2 \times \text{Distance}}{\text{Acceleration}}} $$

Substitute the coordinates of the points A and B to find the distance:

$$ \text{Distance} = \sqrt{(6 - 8)^2 + (5 - 3)^2 + (2 - 8)^2} $$

Then, substitute the values into the time equation:

$$ \text{Time} = \sqrt{\frac{2 \times \text{Distance}}{\frac{7}{4}}} $$

Sebastian Acosta · Answer

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

$$ s = ut + \frac{1}{2}at^2 $$

where:
- $ s $ is the displacement,
- $ u $ is the initial velocity (which is assumed to be zero as the object starts from rest),
- $ a $ is the constant acceleration, and
- $ t $ is the time taken.