math for the physics of a roller coaster
A roller coaster cart of mass m = 339 kg starts
stationary at point A, where h1 = 22.7 m and
a while later is at B, where h2 = 7.1 m.
The acceleration of gravity is 9.8 m/s2 .
What is the speed of the cart at B, ignoring
the effect of friction?
Answer in units of m/s.
sorry it is wrong and the diagram say point A is up and it go down and go up and go down and the last part is B. it is like a W. it start all the way up. the middle curve is shorter than point B. i hope you understand.
The question does not mention the direction of movement of the roller coaster cart. We assume it to be in vertically downward direction.
In this case the the speed of the cart is purely due to acceleration due to gravity (g) as it falls from a height of 22.7 m to 7.1 m.
Thus the distance travelled by the cart is
Distance = s = 22.7 - 7.1 = 15.6
We know that:
s = (gt^2)/2
Where: t = time taken to move the distance s when starting from velocity 0 and accelerating under the influence of g, the acceleration due to gravity.
Substituting values of s and g in above equation:
15.6 = (9.8/2)t^2
t^2 = (15.6*2)/9.8 = 4
t = 4^(1/2) = 2 s
Thus the cart takes 2 seconds to reach the point B.
The speed of the cart is then given by:
Speed = g*Time
= 9.8*2 = 15.6 m/s
Speed of cart at B = 15.6 m/s