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 (3 ,1 ,4 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.

1 Answer
Mar 27, 2018

t=3.24

Explanation:

You can use the formula s=ut+1/2(at^2)
u is initial velocity
s is distance travelled
t is time
a is acceleration

Now, it starts from rest so initial velocity is 0

s=1/2(at^2)

To find s between (6,7,2) and (3,1,4)
We use distance formula
s=sqrt((6-3)^2+(7-1)^2+(2-4)^2)
s=sqrt(9+36+4)
s=7

Acceleration is 4/3 meters per second per second

7=1/2((4/3)t^2)
14*(3/4)=t^2
t=sqrt(10.5)=3.24