- #1

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## Homework Statement

[/B]

Figure 5-55 gives, as a function of time t, the force component

F

_{x}that acts on a 3.00 kg ice block that can move only along

the x axis. At t = 0, the block is moving in the positive direction of

the axis, with a speed of 3.0 m/s.What are its (a) speed and (b) direction

of travel at t = 11 s?

## Homework Equations

[/B]

##\vec{F}=m \vec{a}##

##a_{avg}= \frac{a_1-a_0}{t_1-t_0}##

##V = V_0 + \vec{a}t##

## The Attempt at a Solution

So, if I understand the problem correctly the graph is giving me F

_{x}as a funcion of t, assuming that F

_{x}is the only horizontal force acting on the block a should be given by:

##\vec{F}=m \vec{a}## → ##\vec{a}= \frac {\vec{F}}{m}##

And I think is safe to use the slope formula for the non-horizontal changes in a because they are just a line at a slope, so a

_{avg}should be equal to a

My thought process for the problem is that there is not a "smart way" to solve it, we have to just evaluate V from start to finish in order to get V

_{11}

So, starting with the problem:

V

_{0}= 3 m/s

m = 3 Kg

a

_{0}= 2/3 m/s

^{2}

a

_{2}= 2 m/s

^{2}

a

_{avg0/2}= 2/3 m/s

^{2}

a

_{2/5}= 2 m/s

a

_{avg5/6}= -2 m/s

^{2}

a

_{avg6/9}= -4/3 m/s

^{2}

a

_{avg9/11}= 4/3 m/s

^{2}

Now that we have the change in acceleration for each part of the graph we can simply calculate the final velocity (I'm using unnecessary parenthesis in the formula just to make more clear the different time frames):

V

_{11}= 3 m/s + (2/3 m/s

^{2}⋅ 2s) + (2 m/s ⋅ 3s) + (-2 m/s

^{2}⋅ 1s) + (-4/3 m/s

^{2}⋅ 3s) + (4/3 m/s

^{2}⋅ 2s) = 7 m/s

So the answer should be

(a) s

_{11}= 7 m/s

(b) Positive direction

But the answer in the book (Halliday,Resnik, Walker 9th Ed) is 8 m/s, and I really have no idea what could be wrong in my thought process and calculation, any clue?