Initial Acceleration Homework: Find V0, a, and x0

[Heat]

Homework Statement

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:

Homework Equations

http://physics.webplasma.com/image/page04/kin.gif

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?

 

[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!

 

[Doc Al]

Why guess?

Compare this specific equation:

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

With the general equation for x(t) (the second equation):

http://physics.webplasma.com/image/page04/kin.gif

What must "a" equal?

 

[Heat]

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

 

[G01]

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

Looks good!

 

[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.

 

[Doc Al]

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.

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.)

 

[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.

 

[learningphysics]

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

 

[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?"

 

[Doc Al]

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.)

 

[Heat]

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

 

[Doc Al]

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

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?)

 

[learningphysics]

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.)

Those jerks!

 

[Heat]

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?)

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:

 

[learningphysics]

Yup. That's right.

 

[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.

 

[Doc Al]

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

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

 

[Heat]

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

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?

 

[Doc Al]

Do it! ...

 

[Heat]

here is what I did

-.0625t^2 + 2t + 50 = 60
-.0625t^2 + 2t =10
t(-.0625t + 2) = 10
t = 10s and t= -128s.

So if I am correct the time should be 10second.

I guess not, as 10 seconds is invalid. but it cannot be -128 seconds, as time can't be negative.

 

[learningphysics]

You should also check 40cm.

 

[learningphysics]

here is what I did

-.0625t^2 + 2t + 50 = 60
-.0625t^2 + 2t =10
t(-.0625t + 2) = 10
t = 10s and t= -128s.

How did you do this last step? I don't understand how you solved here..

 

[Doc Al]

here is what I did

-.0625t^2 + 2t + 50 = 60
-.0625t^2 + 2t =10
t(-.0625t + 2) = 10
t = 10s and t= -128s.

This time you need the quadratic formula!

 

[Heat]

I factored t out so it could be

t (-.0625t + 2) = 10

t =10 and -.0625t + 2 = 10

what do you mean by checking 40 cm?
but wouldn't factoring work just as well as quadratic formula?

Let me try quadractic and compare results. :)

 

[Doc Al]

I factored t out so it could be

t (-.0625t + 2) = 10

t =10 and -.0625t + 2 = 10

That dog won't hunt. If the product of two factors is zero, then you can set each factor equal to zero. But not if the product is 10 (or any other non-zero number).

 

[Heat]

Ok after doing quadratic I got the following:

-2 +- sqrt (4-4(-.0625)(-10)) / -.125


-2 +- 1.224744871 / -.125

= 6.20204 seconds and 25.8 seconds

both positive, which do I use?

 

[Doc Al]

Hint...

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

 

[Heat]

sorry about that, I forgot the question, many thanks.

 

[learningphysics]

what do you mean by checking 40 cm?

Ah... my mistake. I thought we should check both 40cm and 60cm, because they're both 10cm from 50cm... but the initial velocity is positive... so it's going to reach 60cm before it reaches 40cm anyway. And since they only need the earliest time, we can ignore the 40cm.