Question

The weight of passengers on a roller coaster increases by $50\%$ as the car goes through a dip with a $30m$ radius of curvature. What is the car’s speed at the bottom of the dip?

Short answer

The speed of the car at the bottom of the dip is$12.12\mathrm{m}/\mathrm{s}$

Step-by-step solution

Given information

Given in the question that, The weight of passengers on a roller coaster increases by $50\%$as the car goes through a dip with a $30m$ radius of curvature.

Explanation

The force acting on the passenger is equal to:

$F=N-mg$

Where,

$N$indicates the normal force

$mg$shows the force of gravity

From the information we observed that, if a person feels like their weight is increased by $50\%$at the bottom that means normal force is $50\%$stronger that the Force of gravity.

Applying the equation of velocity

According to the information the value of $F=0.5mg$

The net force must be equal to the centripetal force.

Therefore,

$0.5mg=\frac{m{V}^2}{R}$

Here,

$m$is the mass, $V$is the velocity and $R$is the radius.

$V^2=0.5gR$

$V=\sqrt{0.5gR}$

$R=30m$

$V=\sqrt{0.5\times 9.8\times 30}$

$V=12.12435$

Final answer

The speed of the car at the bottom of the dip is$12.12\mathrm{m}/\mathrm{s}$