An object is at rest at #(4 ,5 ,1 )# and constantly accelerates at a rate of #4/3 m/s^2# as it moves to point B. If point B is at #(7 ,2 ,6 )#, 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 #(4 ,5   ,1    )# and constantly accelerates at a rate of #4/3  m/s^2# as it moves to point B.  If point B is at #(7  ,2   ,6  )#, how long will it take for the object to reach point B? Assume that all coordinates are in meters.

Nathan Ashcroft · Accepted Answer

It takes #3.14# s.

The equation of motion for a constant accelerating object is #s=1/2at^2# where #s# is the space traveled, #a# is the acceleration and #t# is the time. We are interested in the time that is #t=sqrt((2s)/a)#. The acceleration is given, then we need the space. The distance between two point in space #p_1(x_1, y_1, z_1), p_2(x_2, y_2, z_2)# is given by #d=sqrt((x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2)#. For us #d=sqrt((4-7)^2+(5-2)^2+(1-6)^2)# #=sqrt((-3)^2+(3)^2+(-5)^2)# #=sqrt(9+9+25)=sqrt(43)\approx6.56#. This is the distance that the object has to travel, so we can write #s=6.56# m. We now have all the ingredients #t=sqrt((2s)/a)# #=sqrt((2*6.56" m")/(4/3"m"/"s"^2))# #=sqrt((13.12" m")/("m")*3/4 "s"^2)# #=sqrt(9.84" s"^2)# #=sqrt(9.84" s"^2)\approx3.14" s"#.

Elena Albarran · Answer

undefined To find the time it takes for the object to reach point B, we first need to find the displacement vector from the initial position to point B. Then, we can use the kinematic equation $s = ut + \frac{1}{2}at^2$ to find the time, where $s$ is the displacement vector, $u$ is the initial velocity (which is 0 since the object starts at rest), $a$ is the acceleration, and $t$ is the time.The displacement vector from (4, 5, 1) to (7, 2, 6) is (7-4, 2-5, 6-1) = (3, -3, 5).

The magnitude of this displacement vector is √(3² + (-3)² + 5²) = √(9 + 9 + 25) = √43.

Using the formula $s = ut + \frac{1}{2}at^2$, where $s$ is the magnitude of the displacement vector, $u$ is the initial velocity (0 m/s), $a$ is the acceleration (4/3 m/s²), and $t$ is the time, we can solve for $t$:

$\sqrt{43} = 0 \cdot t + \frac{1}{2} \cdot \frac{4}{3} \cdot t^2$

$\sqrt{43} = \frac{2}{3} t^2$

$t^2 = \frac{3}{2} \cdot \sqrt{43}$

$t = \sqrt{\frac{3}{2} \cdot \sqrt{43}}$