A: Calculate Ferlini's acceleration = 12.3 m/s / 0.16 s = 76.875 m/s^2

[cherryrocket]

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

 

[learningphysics]

1. looks good to me.

For 2, yes that is the right explanation. Acceleration is a change in velocity. Velocity is a magnitude and direction. But speed is just the magnitude of velocity. So if velocity keeps the same magnitude but changes direction... that means that speed remains the same, but velocity changes.

For 3, try to apply the equation d = v1*t + (1/2)at^2

For 4, Ferlini is accelerating from 0 to 12.3m/s over a distance of 2.80m. ie: he is accelerated from rest to 12.3m/s over a distance of 2.80m (which is the length of the cannon) before being fired out of the cannon.

d = v*t can't be used here unless v is average velocity... in other words, the correct equation is:

[tex]d = \frac{(vi + vf)}{2}*t[/tex]

using this equation, you can proceed the same way you did to get the acceleration. ie: get the time, then get acceleration...

Also, there's another displacement equation you can use to directly get acceleration, without solving for time first.

either way of doing the problem is fine. one way is just a little quicker than the other.

 

[cherryrocket]

Confused!

For 3, try to apply the equation d = v1*t + (1/2)at^2

so, what you did was rearrange vi=d-1/2at^2 / t, right?

d= (0m/s)(3.85s) + (1/2)a(3.85s)^2
= 0 + 7.41a
= 7.41a

but I still can't find a because I don't have the average velocity. I'm so confused...

 

[learningphysics]

so, what you did was rearrange vi=d-1/2at^2 / t, right?

d= (0m/s)(3.85s) + (1/2)a(3.85s)^2
= 0 + 7.41a
= 7.41a

but I still can't find a because I don't have the average velocity. I'm so confused...

Haven't you seen the equation d = v_i*t + (1/2)at^2 before? This is one of the equations for displacement with accelerated motion...

there are others also, such as

[tex]d= \frac{(v_i + v_f)}{2}*t[/tex]

and

[tex]v_f^2 = v_i^2 + 2ad[/tex]

if these don't seem familiar be sure to look them over in your textbook...

Also, you do have the acceleration... what is the acceleration of an object that is falling?

 

[cherryrocket]

Yes, I think that perhaps I should read over the whole chapter on Accelerated Motion before trying to do this module. Well, thanks anyway.