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

Expert verified

The speed of the car at the bottom of the dip is12.12m/s

Step by step solution

01

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.

02

Explanation

The force acting on the passenger is equal to:

F=N-mg

Where,

Nindicates the normal force

mgshows 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.

03

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=mV2R

Here,

mis the mass, Vis the velocity and Ris the radius.

V2=0.5gR

V=0.5gR

R=30m

V=0.5×9.8×30

V=12.12435

04

Final answer

The speed of the car at the bottom of the dip is12.12m/s

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