The fast french train known as the tgv (train à grande vitesse) has a scheduled average speed of 216 km/h. (a) if the train goes around a curve at that speed and the acceleration experienced by the passengers is to be limited to 0.060g, what is the smallest radius of curvature for the track that can be tolerated? (b) at what speed must the train go around a curve with a 1.80 km radius to be at the acceleration limit?

a)  r = 6122 m and b) v = 32.5 m / s

Explanation:

a) The train in the curve is subject to centripetal acceleration

         a = v2 / r

Where v is The speed and r the radius of the curve

They indicate that the maximum acceleration of the person is 0.060g,

        a = 0.060 g

        a = 0.060 9.8

        a = 0.588 m /s²

Let's calculate the radius

        v = 216 km / h (1000m / 1km) (1 h / 3600 s =

        v = 60 m / s

        r = v² / a

        r = 60² /0.588

        r = 6122 m

b) Let's calculate the speed, for a radius curve 1.80 km = 1800 m

        v = √a r

        v = √( 0.588 1800)

        v = 32.5 m / s

the simple machine has three main features. they are:

fulcrum is the center point or pivot point of the machine. effort is input force/ applied force. load is external force/ output.

the keisha 's machine labels are

x : load

y: fulcrum

z: lever