## Position Formula :

The position formula the rate of change the displacement when a change in position takes place is Δx is equal to the presuming the golf ball moves from position x_{2}.minus x_{1} in meter. Hence position formula can be written as

**Δ _{x} =x_{2(m)} –x_{(m)}**

The position formula rate change in position at any moment of time X(t)_{(s) } in second is equal to the one to two the product of the acceleration the body possesses α_{ (rad/s2)} in radsecter^{2} multiply the square of body with time is t^{2}_{(s)} in second and addition of velocity of the body v_{0(m/s)} in meter per second and multiply the body with time t_{(s)} in second and addition of initial position of the body a x_{0.}Hence the position formula can be written as

**X(t) _{(s)} = 1/2 *α_{ (rad/s2) }* t^{2}_{(s)} +v_{0(m/s) }*t_{(s)} +x_{0}**

Where,

the position of the body with time t → x(t)

the initial velocity of the body→ v0

the acceleration the body possesses → α in rad/s

the initial position of the body →x_{0}

## Sample Problems :

### Example 1 :

Calculate the position of the person at the end time 4s if the initial velocity of the person is 8m/s and angular acceleration is 3 m/s^{2}?In a^{ }person travels 60m distance.

### Solution :

Known,

V_{0} =8 m/s

X_{0} =60m

α =3 m/s

t= 4s

The change in position of the person at time t is

X(t) = 1/2 α t^{2} +v_{0 }t +x_{0}

X(4)=1/2 ×3 ×(4)^{2}+8 ×4+60

X(4)=24+32+60

X(4)=116m

### Example 2 :

Calculate the position of the person at the end time 3s if the initial velocity of the person is 4m/s and angular acceleration is 5 m/s^{2}? In a person travels 90m distance. .

### Solution :

Known,

V_{0} =4 m/s

X_{0} =90m

α =5m/s

t= 3s

The change in position of the person at time t is

X(t) = 1/2 α t^{2} +v_{0 }t +x_{0}

X(3)=1/2 ×5 ×(3)^{2}+4 ×3+90

X(3)=22.5+102

X(3)=124.5 m