# Position Formula

When something travels from a point to the other, it is called as displacement. Presuming the golf ball moves from position x1 to position x2.

The position change Δx (position formula) is articulated as,

$\Delta&space;_{x}=x_{2}-&space;x_{1}$

Where, the first position of the body is x1,
the second position after undergoing displacement is x2 ,
the rate of change of displacement when change in position takes place is Δx.

If the body changes its position after time t the rate of change in position at any moment of time t, x(t) is articulated as,

$x(t)=\frac{1}{2}\alpha&space;t^{2}+v_{\circ&space;}t+X_{\circ&space;}$

Where,
the position of the body with respect to time t is x(t) ,
the initial velocity of the body is v0 ,
the acceleration the body possesses is α ,
the initial position of the body is x0.

Position Formula Solved Examples

Let’s examine some sample questions on position:

Problem 1: A boy who has an initial velocity of 3 m/s moves for a distance of 20 m. If it’s angular acceleration is 2 m/s2, compute the position of the boy at the end of 5s.

Known :
v0  (Intial velocity)  = 3m/s,

x0  (distance)  = 20 m,

α (angular acceleration)  = 2 m/s2,

t (time) = 5s

The alteration in position of the boy at instant of time t is articulated as

$x(t)=\frac{1}{2}\alpha&space;t^{2}+v_{\circ&space;}t+X_{\circ&space;}$

x(6) = 0.5 × 2 m/s2 × (5s)2+ 3 m/s × 5s +  20 m
= 25m + 15m + 20 m
= 60 m.

Problem  2: A man at a distance of 6m from his home travels some distance to reach his workplace which is 100 m from his home. Calculate the change in position.