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

Layla Ahmed · Accepted Answer

The time taken is #=2.71s#

The following equation of motion will be used:

#s=u_0t+1/2at^2#

#u_0=0#

So,

#s=1/2at^2#

#a=7/4ms^-2#

The distance between 2 points #A=(x_1,y_1,z_1)# and #B=(x_2.y_2.z_2)# is

#=sqrt((x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2)#

Here,

#A=(8,2,5)# and #B=(2,3,7)#

So,

#s=sqrt((2-8)^2+(3-2)^2+(7-5)^2)#

#=sqrt(36+1+4)#

#=sqrt41#

Utilizing the motion equation,

#t^2=(2s)/a#

#t^2=(2*sqrt41)/(7/4)=7.32#

#t=sqrt7.32=2.71s#

Julia Aguilar · Answer

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

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

where $s$ is the displacement, $u$ is the initial velocity, $a$ is the acceleration, and $t$ is time.

Since the object is initially at rest, $u = 0$, the equation simplifies to:

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

Calculate the displacement ($s$) using the distance formula in three dimensions:

$$ s = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2 + (z_B - z_A)^2} $$

Substitute the known values and solve for $t$.