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I'm trying to solve this displacement problem but I dont know where to even start. If you could please explain I would really appreciate it. My professor didn't teach and refuses to help.

The velocity (in meters/sec) of a particle moving along a straight line is given by $ v\, ( t ) =1 t^2 -4 t + 4 $, where $ t $ is measured in seconds. Answer the following questions given that the initial position$ \, \,s (0) =1 $.

What is the meter position of the particle at any given time $ t $ ?

What is the meter position of the body at time $ t=4 $?

What is the meter position of the particle at time $ t=7 $ ?

What is the displacement of the particle on the time interval $ 4 \leq t \leq 7 $ ?

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For the motion of a particle we have: $$\boldsymbol a(t) = \frac{d\boldsymbol v(t)}{dt},~~~~~~~~\boldsymbol v(t) = \frac{d\boldsymbol s(t)}{dt}.$$ Now you should be able to find $s(t)$ and answer the questions.

Warning! Think about the last question: is the displacement $s(7)-s(4)$? Or is it the total amount of meters the particle travelled? Analogously: If I go from Washington to Buenos Aires, and then to Miami, was my displacement the distance between Washington and Miami? or was it the distance between Washington and Buenos Aires plus the distance between Buenos Aires and Miami?

Recall also that (this comes from Physics and from integrating $a=\frac{dv}{dt}$):

$$\boldsymbol s(t) = \boldsymbol s(0)+\boldsymbol v(0)t+\frac{1}{2}\boldsymbol a(t)t^2$$

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  • $\begingroup$ so when you take der 2t-4+4 what to do next? substitute the values? Thank Vladimir $\endgroup$ – Ozi Nov 8 '14 at 0:29
  • $\begingroup$ Thanks, could you please explain what is a? $\endgroup$ – Ozi Nov 8 '14 at 0:41
  • $\begingroup$ $a$ stands for acceleration. $a=\dfrac{\text{Change in velocity}}{\text{Change in Time}}=\dfrac{\Delta v}{\Delta t}$. $\endgroup$ – Vladimir Vargas Nov 8 '14 at 0:43
  • $\begingroup$ @VladimirVargas "When you take the derivative of a function that describes the velocity of an object, you will obtain a function that describes the position of an object." This is plain wrong. Derivative of velocity gives acceleration. $\endgroup$ – G-man Nov 8 '14 at 9:34
  • $\begingroup$ Well what to do next after taking der? $\endgroup$ – Ozi Nov 8 '14 at 23:10

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