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I need help on this problem, please if someone could look at it and help me out

  • Thread starter miranda82
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  • #1
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I need help on this problem, please if someone could look at it and help me out!!

Suppose you measured the velocity of an air cart twice as it moves along an air track. At T1= 0.2 Sec the velocity is V1= 0.4 m/s.
At T2= 0.6 Sec the velocity is V2= 0.9 m/s.
What is the average acceleration of the cart?
 

Answers and Replies

  • #2
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.4/.5 is the difference. if you want acceleration per second, multiple .4 and .5 by 2.5. you should get A = 1.25 m/s. i think lol.
 
  • #3
Doc Al
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What's the definition of acceleration?
 
  • #4
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the change in velocity over time.
 
  • #5
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Average aceleration = Change Velocity / Time
 
  • #6
Doc Al
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miranda82 said:
What is the average acceleration of the cart?
If you understand the definition of acceleration, you should be able to figure out the answer. (And you should also be able to tell whether Gecko's answer is correct. :smile: )
 
  • #7
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i think you are correct, seeing that the formula says that, and it is last minus first.
 
  • #8
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Thank you every one
 
  • #9
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I worked this out but i need advice if i did it correct!!
It goes along with the last problem i did!

If the acceleration is due to gravity what is the angle of inclination of the air track?

I did Average acceleration = /G/ SIN Degree

1.25 m/s= /9.8/ SIN Degree

1/sin (9.8)/ (1.25) = -2.18

Is this way off ??????
 
  • #10
Doc Al
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The relationship between the acceleration and the angle of incline is correct, but your calculation is not.
[itex]a = g sin\theta[/itex], so [itex]sin\theta = a/g = 1.25/9.8 = 0.128[/itex]. To find the angle, use your calculator to take the inverse sine. (I'll leave that to you.)

Note: The units of acceleration are [itex]m/s^2[/itex], not [itex]m/s[/itex].
 

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