Acceleration,velocity and time of a car

[skp123]

Homework Statement

A 20 kg car slides 5 m down the smooth(frictionless) ramp with an initial celocity 0.

a) Determine the acceleration when the car moves from A to B.
b)Determine the velocity of the car in point B.
c)Determine the point (distance L) where it strikes the ground at C.
d) How long does it take from A to C ?Image of the car

http://img205.imageshack.us/img205/1/carcx.jpg

Homework Equations

The Attempt at a Solution

Can I find the acceleration doing this.

G=mgsin(30) = 10 N

F=ma -> a=F/m -> a=2 m/s

If this is wrong , can anybody tell what i am doing wrong.I haven't solved problems like this before and I want to know if i am doing it right or no

 

[jegues]

You've defined you x-axis parallel to the plane and your y-axis perpendicular to the plane so the way you've solved for the acceleration seems right to me.

See if you can solve b. You know the acceleration, the intial velocity and the distance he travels... If we could solve for the time it takes him to get to point b, maybe we could calculate his final velocity ;)

Start with that and if you can't solve c, and d let us know where you're having troubles.

 

[skp123]

Thank you. Now b -> we know that a=vdv/ds -> integrate and we find v=2\sqrt{5}.
Now i have problems finding the distance L. If the angle between L and C is also 30 degree i can find it but i don't think it is 30 degree. And i also don't think that BC is a straight line. So can you help me finding this distance L ?

 

[jegues]

The velocity we already defined was parallel to the plane, so we need to break this down into horizontal and vertical vectors and look at that way in order to solve the distance L. See the figure attached. You can solve for the horizontal component of velocity and the vertical component of velocity right? How can you use this new information with the information we know from the drawing ( h = 1 ) to solve the distance L.

Note that the horizontal velocity will have no acceleration throughout its free fall only the vertical velocity is changing ;)

 

[skp123]

No, I can't. Sorry but I need more help. I am trying to understand it but i can't.

Now we need to find v_y - right ? We have v=2\sqrt{5} , we have the angle -> v_y = vsin(30) -> v_y = \sqrt{5} .

 

[jegues]

DISREGARD THIS POST, I misread your post above.

 

[jegues]

Okay so you've established the intial velocity of that segment of its motion. We know the distance it travels vertically so what formula can we use to solve for the time it will take to reach the ground?

 

[skp123]

Ok, but first we need to find this distance L. And I can't find it

 

[jegues]

You need to find what I mentioned above in order to solve for L :p

 

[skp123]

OK. So t=S/v -> t=5 / 2\sqrt{5} -> t=\sqrt{5}/2 . But this is the time from A to B. Now what should i do ?

 

[jegues]

You're not looking for the time from A to B, you're looking for the time from B to the C.

 

[skp123]

OK.So . Can I use this formila t=s/v ? If yes -> t=\sqrt{5} / 10 . And not let's find distance L . S=vt -> 2\sqrt{5} * \sqrt{5}/10 = 1. Therefore distance L= 1

 

[jegues]

You are using the Vy to help you find t, you know you pass through a vertical distance of 1m so the formula you're looking to use is as follows:

h = Voyt + 1/2at^2, solve for t

Hint: Quadratic formula

 

[skp123]

OK, but where this formula come from. Whats the origin formula. Sorry for my questions , but i just can't get it.

 

[jegues]

D= Vot + 1/2at^2 is a standard formula for constant acceleration problems.

 

[skp123]

Ok. I think i don't know this formula . Thank you very much