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 #(6 ,5 ,1 )#, 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 #(6  ,5   ,1  )#, how long will it take for the object to reach point B? Assume that all coordinates are in meters.

Adrian Ayala · Accepted Answer

time #t=2.480821248" "#seconds

Given object at rest #v_0=0" "#initial velocity distance traveled by the object #s=sqrt((x_B-x_A)^2+(y_B-y_A)^2+(z_B-z_A)^2)# #s=sqrt((6-8)^2+(5-2)^2+(1-5)^2)# #s=sqrt(4+9+16)# #s=sqrt(29)# the formula to compute t #s=v_0t+1/2at^2# #sqrt(29)=(0)*t+1/2(7/4)t^2# #t^2=8/7sqrt29# #t=(8/7sqrt29)^(1/2)# #t=2.480821248" "#seconds God bless....I hope the explanation useful

Adrian Ayala · Answer

undefined To find the time (t), you can use the kinematic equation:

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

where:
- $ s $ is the displacement,
- $ u $ is the initial velocity (which is 0 since the object is at rest),
- $ a $ is the acceleration,
- $ t $ is the time.

Given that $ u = 0 $, $ a = \frac{7}{4} $, and the displacement $ s $ is the distance between the initial and final positions (points A and B), you can use the coordinates to calculate $ s $ in each dimension (x, y, z).

Finally, solve for $ t $ using the calculated values for $ s $ and $ a $.