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france has claimed the worlds fastest train. A modified TGV can accelerate to a maximum speed of 575 km/h. If the train started from rest and reached its maximum velocity of 575 km/h at 8.00 km, what would the magnitude of the acceleration of the train be?

 

5 years ago

Answered By Wendy W

speed = acceleration*time, so time = speed/acc. = 575/acc.

distance = 1/2*acceleration*time2

therefore, 8 = 1/2*acc.*(575/acc.)= 1/2*5752/acc.

magnitude of acc. = 20664.1km/h2


5 years ago

Answered By Andres C

Just note what you have and what they're asking. they give you the initial speed (0km/h) and the final speed (575km/h), as well as the distance (8km). What they're asking is for the acceleration. note that out of all the kinematics equations (which should be given to you in your formula sheet) the only one that has all the information you have is Vf^2 = Vi^2 + 2ad

when you plug everything in then a = 2.07*10^5 km/h or 5740m/s


5 years ago

Answered By Clifton P

First we should put all the pieces into standard units

d=8km=8000m

Vf = 575 km/h = 159.7m/s

(Easy conversion of km/h to m/s is

1m/s = 3.6km/h)

using the equation

Vf^2 = Vi^2 + 2ad

and noting that it starts from rest (Vi=0)

we can solve for a

(Vf^2 - 0) / 2d = a

(159.7^2) / 16000 = a

25504.1 / 16000 = 1.59m/s^2 = a

so the train accelerates at

~1.6m/s^2