Assignment 1 - Handwritten Questions (Physics)

A particle moves along the *x axis* according to the equation\ [**x=** **2.00** **+** **3.00t**  **−** **1.00t**^**2**^]{.math.inline}**,** where *x* is in meters and *t* is in seconds. At ***t* = 3.00 s**, find (a) the position of the particle, (b) its velocity, and (c) its acceleration. a)- posit...

A particle moves along the *x axis* according to the equation\ [**x=** **2.00** **+** **3.00t**  **−** **1.00t**^**2**^]{.math.inline}**,** where *x* is in meters and *t* is in seconds. At ***t* = 3.00 s**, find (a) the position of the particle, (b) its velocity, and (c) its acceleration. a)- position at t=3.00s x=2.00 + 3.00t -- 1.00[*t*2]{.math.inline} x(3.00s)= 2.00 + ([3.00 × 3.00)]{.math.inline} ­-- ([1.00 × 3.00^2^)]{.math.inline} = 2.00m x=2.00m b)- velocity at t=3.00s v =[\$\\ \\frac{\\text{dx}}{\\text{dt}}\$]{.math.inline} = 3.00 + --2.00t v(3.00s)= 3.00 + [(--2.00×3.00)]{.math.inline} = --3.00 m/s c)-acceleration at t=3.00s a= [\$\\frac{\\text{dv}}{\\text{dx}}\$]{.math.inline} = --2.00 a(3.00s)= --2.00 [*m*/*s*^2^]{.math.inline} 2. The height of a helicopter above the ground is given by [**h** **=** **3.00t**^**3**^]{.math.inline}**,** where *h* is in meters and *t* is in seconds. At **t = 2.00 s**, the helicopter releases a small mailbag. How long after its release does the mailbag reach the ground? t after the mailbag release to the ground=?? h(2.00s)= [3.00 × 2.00^3^]{.math.inline} = 24.00 m h=[*v*~*i*~]{.math.inline}t + [\$\\frac{1}{2}at\^{2}\$]{.math.inline} h= 24.00 m , [*v*~*i*~]{.math.inline}= 0 24.00= 0 + [\$\\frac{1}{2} \\times 9.8 \\times t\^{2}\$]{.math.inline} 24.00= 4.9[*t*^2^]{.math.inline} [\$\\frac{24.00}{4.9}\$]{.math.inline}=[\$\\frac{4.9}{4.9} \\times t\^{2}\$]{.math.inline} 4.9=[*t*^2^]{.math.inline} [\$\\sqrt{4.9}\$]{.math.inline} = [\$\\sqrt{t}\$]{.math.inline} t= 2.2s