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

1. Oct 19, 2004

### miranda82

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?

2. Oct 19, 2004

### Gecko

.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. Oct 19, 2004

### Staff: Mentor

What's the definition of acceleration?

4. Oct 19, 2004

### Gecko

the change in velocity over time.

5. Oct 19, 2004

### miranda82

Average aceleration = Change Velocity / Time

6. Oct 19, 2004

### Staff: Mentor

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. )

7. Oct 19, 2004

### miranda82

i think you are correct, seeing that the formula says that, and it is last minus first.

8. Oct 19, 2004

### miranda82

Thank you every one

9. Oct 19, 2004

### miranda82

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. Oct 19, 2004

### Staff: Mentor

The relationship between the acceleration and the angle of incline is correct, but your calculation is not.
$a = g sin\theta$, so $sin\theta = a/g = 1.25/9.8 = 0.128$. To find the angle, use your calculator to take the inverse sine. (I'll leave that to you.)

Note: The units of acceleration are $m/s^2$, not $m/s$.