# Initial Acceleration

1. Sep 9, 2007

### Heat

1. The problem statement, all variables and given/known data

A truck drives along a straight line, which we will call the x-axis with the positive direction to the right. The equation for the turtle's position as a function of time is:

x(t) = 50.0cm + (2.00cm/s)t - (.0625cm/s^2)t^2

Find initial velocity:
Find initial acceleration:
Find initial position:

2. Relevant equations
http://physics.webplasma.com/image/page04/kin.gif [Broken]

3. The attempt at a solution

I found that initial velocity is 2.00cm/s, as the equation states that V0 is initial velocity.

I found initial position, by knowing that if time was zero, then the obvious would be to have it at 50cm.

The acceleration is giving me the problem, I would think that it was .0625cm/s^2, but it is wrong. Then I would guess that since the initial velocity is 2.00cm/s then, acceleration at that point in time might be the same...is this right?

I found this equation for instantaneous acceleration in which a= delta vx / delta t.

So this would be

2.00cm/s / 0 = 0 acceleration?

Last edited by a moderator: May 3, 2017
2. Sep 9, 2007

### G01

What about the 1/2 in this equation?:

$$x(t)=x_0+v_0t+1/2at^2$$

Also, remember that you have -.0625, not +.0625

EDIT: Perfect Timing Doc Al!

Last edited: Sep 9, 2007
3. Sep 9, 2007

### Staff: Mentor

Why guess?

Compare this specific equation:
With the general equation for x(t) (the second equation):

http://physics.webplasma.com/image/page04/kin.gif [Broken]

What must "a" equal?

Last edited by a moderator: May 3, 2017
4. Sep 9, 2007

### Heat

so initial acceleration must equal -.125 cm/s^2

Last edited: Sep 9, 2007
5. Sep 9, 2007

Looks good!

6. Sep 9, 2007

### Heat

Ok, I understood that part, this is another part of the question:

How long after starting does it take the truck to return to its starting point?

I managed to find at what time the velocity of the truck would be zero, and it would be at 16s.

Now I would setup it up like this,

0 = 50.0cm + (2.00cm/s)t - (.0625 cm/s^2)t2

-.0625t^2 + 2t + 50 = 0 <----I would solve for t using quadratic equation, although , I don't believe this is possible since they are cm/s and cm.s^2 etc.

7. Sep 9, 2007

### Staff: Mentor

Looks good to me--solve it! The units for time will be seconds--no problem. (Note that each term in that equation for position has units of length--so everything is perfectly consistent.)

8. Sep 9, 2007

### Heat

Ok, this is what I got

-2 +- sqrt (4-4(-.0625)(50)) / 2(-.0625)

= -2+- 4.06 / -0.125

= -16.48 and 48.48 seconds.

since time cannot be negative, I answered 48.48 seconds, but I got it wrong.

Can you help me find my mistake?

Unless I add 16s (from when velocity was 0 at time 16s) with 48.48seconds, when the truck reaches same point ?

I rechecked my calculation , updated it as above, and still got it wrong.

Last edited: Sep 9, 2007
9. Sep 9, 2007

### learningphysics

Don't round that discriminant to 4.06. Carry everything as much as you can.

10. Sep 9, 2007

### Heat

-2 + - 4.062019202 / -.125 = -16.496 and 48.496

does this seem right?

sould I add the 16s when velocity was 0?

would the answer the 64.496? or 48.496?

A reminder of the question : "How long after starting does it take the truck to return to its starting point?"

11. Sep 9, 2007

### Staff: Mentor

Oops... looks like they tricked us. It says return to the starting point. Assuming it starts at t = 0, the starting position is not x = 0. (You won't even need a quadratic to solve it now.)

12. Sep 9, 2007

### Heat

but if t = 0, then how could be solve for t? I am confused on what to use now :(

13. Sep 9, 2007

### Staff: Mentor

t = 0 will give you the starting position; plug that position into your equation for x(t) and solve for t. (You know that one answer is t = 0; what's the other?)

14. Sep 9, 2007

### learningphysics

Those jerks!!!

15. Sep 9, 2007

### Heat

x(0) = 50.0cm + 2.00 cm/s (t) - 0.0625cm/s^2 (t)^2

x(0) = 50.0cm + 2.00 cm/s (0) - 0.0625cm/s^2 (0)^2

= 50.0cm

50.0cm = 50.0cm + 2.00 cm/s (t) - 0.0625cm/s^2 (t)^2

-.0625t^2 + 2t = 0

t(-.0625t + 2) = 0
t= 0 <---which we know
and

-.0625t +2 = 0

t= 32seconds.

so the time it will take to return to the starting point would be 32seconds. Am i right? :zzz:

16. Sep 9, 2007

### learningphysics

Yup. That's right.

17. Sep 9, 2007

### Heat

Sorry if I am bombarding you guys with questions.

The next one got me too:

At what time is the truck first time a distance of 10.0 cm from its starting point?

This is what I did, assuming that it follows this format:

10 = 50.0 2t - .0625t^2.

t= -40s and 672 seconds.

both are incorrect.

18. Sep 9, 2007

### Staff: Mentor

Lets not get fooled again! The starting point is not at x=0! So what position are we talking about?

19. Sep 9, 2007

### Heat

well if the initial starting point is 50cm, then if I add 10cm more it would be ay 60cm, which I would have to solve for time.

Do I solve this again for 60cm?

20. Sep 9, 2007

### Staff: Mentor

Do it! ...

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