At A, coaster is only associated with potential energy.
At B, coaster is associated with kinetic as well as potential energy.
Since the track is frictionless, no energy will be lost when coaster reaches from point A to point B. Therefore, according to conservation of energy, total energy at A should be equal to total energy at B.
Total energy at A = mgh = mg(12)
Total energy at B = mgh+ mv²/2 = mg(2) + mv²/2
∴12mg = 2mg + mv²/2
∴(12g-2g)×2 = v²
∴v² = 20g
∴v = 14m/s.
Again conserving energy at points B and C.
Total energy at B = 2mg + m(14)²/2
Total energy at C = 4mg + mv²/2
∴2mg + m(14²)/2 = 4mg + mv²/2
Solving this you get,
v = 12.52 m/s.
Therefore, speed of roller coaster at point C is 12.52 m/s.
Answer:
The answer to your question is: t = 2.5 s
Explanation:
Data
vo = 30 m/s
a = -12 m/s2
t = ?
vf = 0 m/s
Formula
vf = vo + at
Substitution
0 m/s = 30 + (-12)t
Solve it for t -30 = -12t
t = -30 / -12
t = 30/12 = 15/6 = 5/2
t = 2.5 s
Answer:
all of above I think sorry if I'm not right but I tried
