## Find Average Velocity of cart rolling down the slope.

In the diagram below, a cart is released from rest at a distance x1 from the first of two photogates. The cart then rolls down the ramp through both photogates. The distance between the photogates is x2. For this problem, you may assume that the cart is frictionless and the acceleration of the cart a=g sin THETA (in degrees).

Question) If x1 = 20 cm, x2 = 50 cm, and THETA = 20 degrees, what is the average velocity of the cart between the photogates (over the distance x2)?

(Here is the link to the image: http://www.facebook.com/photo.php?pi...1&id=645560319 )

This is what I did to solve this problem:

v^2=v0^2 + 2ax

a=-9.8sin20degrees = 3.352

x1=20
x2=50
v0x=0

vx1^2 = 0 + 2(3.352)(20),
vx1^2= SQUARE ROOT of 134.072
= 11.5789 cm/s.

vx2^2 = 11.5789^2 + 2(3.352)(50),
vx2=SQUARE ROOT of 469.271
= 21.663 cm/s

Is this answer correct? I seriously doubt if I am using the correct equations. I can't talk to my professor until the day I have to turn my home work in, and I can't find any examples on google :) I will appreciate if somebody will confirm or correct this ASAP! :)

Thankyou,
Arshad_Physic

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 Mentor Blog Entries: 10 Watch the units, you are mixing m and cm together here. Pick one or the other and be consistent. Do you want to use cm? Then g is not 9.8 cm/s^2. p.s. welcome to Physics Forums.
 The initial velocity is zero. The total displacement is 30 cm or 0.3 m, yes? So just use the same kinematic equation you have to find the final velocity using delta x = 30 cm. Then divide this by two.... since the initial velocity is zero. Objects that accelerate at a uniform rate, as in your problem, allow you to take a nice clean average of two velocities. The initial which is zero, and the final velocity which you are going to find now... sorry for the intrusion.

## Find Average Velocity of cart rolling down the slope.

Thanks Redbelly and pgardn! :)

And yeah, my Physics 2425 course forced me to join this forum! haha - But I am glad because this place is AMAZING!!! :)

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