- #1
cherryrocket
- 19
- 0
1. Q: "John slides down on a hill on his toboggan. He starts from rest and speeds up to 18.0 km/h in 8.0 s. Find his acceleration in m/s^2."
A: Convert 18.0 km/h to m/s^2 = 18.0 km/h(0.2778) = 5.00 m/s
a=v/t = 5.00m/s / 8.0s = 0.63 m/s^2
2.Q: Explain how you can accelerate without speeding up.
A: By changing direction when you accelerate. Ex. turning around a corner (I'm not really getting this explanation)
3. An object falls off a cliff and lands 3.85 s later. How high is the cliff?
A: I don't know how to get it, there is only on number given, I don't have any other numbers to put in the average acceleration equation.
4.Q: "Ferlini the human cannonball is shot out of a 2.80 m long cannon with a velocity of 12.3 m/s. Find his acceleration."
A: Is 2.80 m the distance? I don't think it is because it is just what he was shot out off, not the distance covered.
But this is what I did anyway:
t= d/v = 2.0m/12.3m/s = 0.16s = 0.2s
Then I looked for the acceleration:
a=v/t =12.3m/s / 0.2s = 61.5 m/s^2
A: Convert 18.0 km/h to m/s^2 = 18.0 km/h(0.2778) = 5.00 m/s
a=v/t = 5.00m/s / 8.0s = 0.63 m/s^2
2.Q: Explain how you can accelerate without speeding up.
A: By changing direction when you accelerate. Ex. turning around a corner (I'm not really getting this explanation)
3. An object falls off a cliff and lands 3.85 s later. How high is the cliff?
A: I don't know how to get it, there is only on number given, I don't have any other numbers to put in the average acceleration equation.
4.Q: "Ferlini the human cannonball is shot out of a 2.80 m long cannon with a velocity of 12.3 m/s. Find his acceleration."
A: Is 2.80 m the distance? I don't think it is because it is just what he was shot out off, not the distance covered.
But this is what I did anyway:
t= d/v = 2.0m/12.3m/s = 0.16s = 0.2s
Then I looked for the acceleration:
a=v/t =12.3m/s / 0.2s = 61.5 m/s^2