An object is at rest at #(1 ,3 ,5 )# and constantly accelerates at a rate of #4/3 m/s^2# as it moves to point B. If point B is at #(3 ,6 ,4 )#, 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 #(1 ,3   ,5    )# and constantly accelerates at a rate of #4/3  m/s^2# as it moves to point B.  If point B is at #(3  ,6   ,4  )#, how long will it take for the object to reach point B? Assume that all coordinates are in meters.

Elena Albarran · Accepted Answer

The time is #=2.37s#

The distance between the points #A=(x_A,y_A,z_A)# and the point #B=(x_B,y_B,z_B)# is

#AB=sqrt((x_B-x_A)^2+(y_B-y_A)^2+(z_B-z_A)^2)#

#d=AB= sqrt((3-1)^2+(6-3)^2+(4-5)^2)#

#=sqrt(2^2+3^2+1^2)#

#=sqrt(4+9+1)#

#=sqrt14#

#=3.74m#

We utilize the equation of motion.

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

#u=0#

thus,

#d=1/2at^2#

#a=4/3ms^-2#

#t^2=(2d)/a=(2*3.74)/(4/3)#

#=5.61s^2#

#t=sqrt(5.61)=2.37s#

Olivia Askew · Answer

undefined Use the kinematic equation: $ \Delta x = v_0 t + \frac{1}{2} a t^2 $ where $ \Delta x $ is the displacement, $ v_0 $ is the initial velocity, $ a $ is the acceleration, and $ t $ is time.

Given:
Initial position ($ x_0, y_0, z_0 $) = (1, 3, 5)
Final position ($ x, y, z $) = (3, 6, 4)
Acceleration ($ a $) = 4/3 m/s²

Calculate displacement:
$\Delta x = x - x_0$, $\Delta y = y - y_0$, $\Delta z = z - z_0$

Use the displacement values in the kinematic equation to solve for time ($ t $).