An object is at rest at #(6 ,7 ,2 )# and constantly accelerates at a rate of #4/3 m/s^2# as it moves to point B. If point B is at #(8 ,1 ,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 #(6 ,7   ,2    )# and constantly accelerates at a rate of #4/3  m/s^2# as it moves to point B.  If point B is at #(8  ,1   ,7  )#, how long will it take for the object to reach point B? Assume that all coordinates are in meters.

Julia Aguilar · Accepted Answer

#~~3.46s# Here displacement = linear distance between the given points so displacement #s=sqrt((6-8)^2+(7-1)^2+(2-7)^2)=sqrt65m# Initial velocity #u=0# Acceleration #a=4/3ms^-2# time taken t=? So #s=ut+1/2at^2# #sqrt65=0*t+1/2*4/3t^2# #t^2=3/2*sqrt65~~12# #t ~~sqrt12~~3.46s#

Charles Aeschliman · Answer

undefined To find the time it takes for the object to reach point B, we can use the formula for displacement under constant acceleration:

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

where:
- $ \Delta x $ is the displacement (change in position),
- $ v_i $ is the initial velocity (which is 0 since the object is at rest),
- $ a $ is the acceleration (given as $ \frac{4}{3} \, \text{m/s}^2 $),
- $ t $ is the time.

Given that the object starts at position (6, 7, 2) and ends at position (8, 1, 7), we need to find the displacement along each axis (x, y, z):

- Along x-axis: $ \Delta x = 8 - 6 = 2 $ meters
- Along y-axis: $ \Delta y = 1 - 7 = -6 $ meters (note the negative sign indicates the direction)
- Along z-axis: $ \Delta z = 7 - 2 = 5 $ meters

Since the object starts at rest, its initial velocity ($ v_i $) is 0.

Now, we can plug the values into the formula and solve for $ t $:

$$ 2 = 0 \times t + \frac{1}{2} \times \frac{4}{3} \times t^2 $$

$$ 2 = \frac{2}{3} t^2 $$

$$ t^2 = \frac{3}{2} \times 2 $$

$$ t^2 = 3 $$

$$ t = \sqrt{3} $$

Thus, it will take $ \sqrt{3} $ seconds for the object to reach point B.