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 (2 ,3 ,7 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.

1 Answer
Mar 18, 2017

The time taken is =2.71s

Explanation:

We are going to apply the following equation of motion :

s=u_0t+1/2at^2

u_0=0

So,

s=1/2at^2

a=7/4ms^-2

The distance between 2 points A=(x_1,y_1,z_1) and B=(x_2.y_2.z_2) is

=sqrt((x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2)

Here,

A=(8,2,5) and B=(2,3,7)

So,

s=sqrt((2-8)^2+(3-2)^2+(7-5)^2)

=sqrt(36+1+4)

=sqrt41

From the equation of motion,

t^2=(2s)/a

t^2=(2*sqrt41)/(7/4)=7.32

t=sqrt7.32=2.71s