An object is at rest at #(7 ,8 ,4 )# and constantly accelerates at a rate of #5/4 m/s^2# as it moves to point B. If point B is at #(1 ,5 ,3 )#, 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 #(7  ,8   ,4    )# and constantly accelerates at a rate of #5/4  m/s^2# as it moves to point B.  If point B is at #(1   ,5   ,3  )#, how long will it take for the object to reach point B? Assume that all coordinates are in meters.

Sebastian Acosta · Accepted Answer

It takes 3.29 seconds.

First of all we need to know how much distance the objects has to travel. The distance between two points is given by the Pitagora's theorem in 3 dimensions #d=sqrt((x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2)# in our case #d=sqrt((7-1)^2+(8-5)^2+(4-3)^2)=sqrt(36+9+1)=sqrt(46)\approx6.78m# Now that we know that our body will travel for 6.78 m we can use the equation of motion for an accelerating body that is #x=1/2at^2# where #x# is the traveled distance, for us 6.78, #a# is the acceleration, for us #5/4m/s^2# and #t# is the time that we want to find. I plug the values in the equation #6.78=1/2*5/4*t^2# #6.78=5/8t^2# I am interested in #t# so I multiply both sides for #8/5# and do the square root #sqrt(6.78*8/5)=t# #t=sqrt(10.848)\approx 3.29# s.

Nicholas Agrusa · Answer

undefined Using the kinematic equation for uniformly accelerated motion:

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

$$ \Delta y = v_{yi} t + \frac{1}{2} a t^2 $$

$$ \Delta z = v_{zi} t + \frac{1}{2} a t^2 $$

Where:
$ \Delta x = x_B - x_A = 1 - 7 = -6 $ m,
$ \Delta y = y_B - y_A = 5 - 8 = -3 $ m,
$ \Delta z = z_B - z_A = 3 - 4 = -1 $ m,
$ v_{xi} = 0 $ m/s, $ v_{yi} = 0 $ m/s, $ v_{zi} = 0 $ m/s,
$ a = \frac{5}{4} $ m/s².

Substituting the values into the equations:

$$ -6 = 0 + \frac{1}{2} \left( \frac{5}{4} \right) t^2 $$

$$ -3 = 0 + \frac{1}{2} \left( \frac{5}{4} \right) t^2 $$

$$ -1 = 0 + \frac{1}{2} \left( \frac{5}{4} \right) t^2 $$

Solving for $ t $:

$$ -6 = \frac{5}{8} t^2 $$
$$ -3 = \frac{5}{8} t^2 $$
$$ -1 = \frac{5}{8} t^2 $$

$$ t^2 = \frac{-6 \times 8}{5} = -\frac{48}{5} $$
$$ t^2 = \frac{-3 \times 8}{5} = -\frac{24}{5} $$
$$ t^2 = \frac{-1 \times 8}{5} = -\frac{8}{5} $$

Since time cannot be negative, the object will never reach point B. There may be an error in the calculations or assumptions.