Question

An amusement park ride moves a rider at a constant speed of 14 meters per second in a horizontal circular path of radius 10. meters. What is the rider's centripetal acceleration in terms of g, the acceleration due to gravity?

A) 1g
B) 2g
C) 3g
D) 0g

Answer 1

Answer:

B) 2g

Explanation:

Given the following data;

Velocity, v = 14m/s

Radius, r = 10m

To find the centripetal acceleration;

$Acceleration, a = \frac {v^{2}}{r}$

Substituting into the equation, we have;

$Acceleration, a = \frac {14^{2}}{10}$

$Acceleration, a = \frac {196}{10}$

Acceleration, a = 19.6m/s²

In terms of acceleration due to gravity, g = 9.8m/s²

We would divide by g;

Acceleration, a = 19.6/9.8 = 2

Hence, centripetal acceleration = 2g

Therefore, the rider's centripetal acceleration in terms of g, the acceleration due to gravity is 2g.

Answer 2

The rider's centripetal acceleration in terms of g the acceleration due to gravity is 2g.

Centripetal acceleration is the acceleration of a body experiencing circular motion.

Centripetal acceleration is given by:

a = v²/r;

where a is the centripetal acceleration, v is the velocity and r is the radius of the circular path.

Given that v = 14 m/s, r = 10 m, hence:

a = v²/r = 14²/10 = 19.6 m/s²

g = 9.8 m/s²

Hence; a = 19.6 m/s² = 2(9.8) = 2g

The rider's centripetal acceleration in terms of g the acceleration due to gravity is 2g.

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