The velocity of the trainee is **29 m/s or 0.42 rev/s**

### Further explanation

**Acceleration** is rate of change of velocity.

[tex]\large {\boxed {a = \frac{v - u}{t} } }[/tex]

[tex]\large {\boxed {d = \frac{v + u}{2}~t } }[/tex]

*a = acceleration (m / s²)v = final velocity (m / s)*

*u = initial velocity (m / s)*

*t = time taken (s)*

*d = distance (m)*

**Centripetal Acceleration** of circular motion could be calculated using following formula:

[tex]\large {\boxed {a_s = v^2 / R} }[/tex]

*a = centripetal acceleration ( m/s² )*

*v = velocity ( m/s )*

*R = radius of circle ( m )*

Let us now tackle the problem!

__Given:__

Radius of horizontal circle = R = 11.0 m

Force Felt by the Trainee = F = 7.80w

__Unknown:__

Velocity of Rotation = v = ?

__Solution:__

[tex]F = ma[/tex]

[tex]F = m\frac{v^2}{R}[/tex]

[tex]7.80w = m\frac{v^2}{R}[/tex]

[tex]7.80mg = m\frac{v^2}{R}[/tex]

[tex]7.80g = \frac{v^2}{R}[/tex]

[tex]7.80 \times 9.8 = \frac{v^2}{11.0}[/tex]

[tex]v^2 = 840.84[/tex]

[tex]v \approx 29 ~m/s[/tex]

[tex]\omega = \frac{v}{R}[/tex] → in rad/s

[tex]\omega = \frac{v}{2 \pi R}[/tex] → in rev/s

[tex]\omega = \frac{29}{2 \pi \times 11.0}[/tex]

[tex]\omega \approx 0.42 ~ rev/s[/tex]

### Answer details

**Grade: **High School

**Subject:** Physics

**Chapter:** Circular Motion

**Keywords:** Velocity , Driver , Car , Deceleration , Acceleration , Obstacle , Speed , Time , Rate , Circular , Ball , Centripetal