Question

A particle has such a curvilinear motion that its x coordinate is defined by 𝑥 = 5𝑡 3 −105𝑡 where 𝑥 is in inches and 𝑡 is in seconds. When 𝑡 = 2 𝑠, the total acceleration is 75 𝑖𝑛/𝑠 2 . If the y component of acceleration is constant and the particle starts from rest at the origin when 𝑡 = 0, determine its total velocity when 𝑡 = 4 𝑠.

Answer: 𝒗 = 𝟏𝟖. 𝟕𝟓 𝒇𝒕/s

Answer 1

$\begin{aligned} &x = 5t³ - 105t\\ \\ &\dfrac{dx}{dt} = 5t² - 105\\ \\ &\therefore a_x(t) =\dfrac{d²x}{dt²} = 10t \end{aligned}$

$a_x(2) = 10(2) = 20 \ in/s²$

$\textsf{Since, } \sqrt{(a_x)² +(a_y)²} = a$

$75 = \sqrt{20² + (a_y)²}$

$5625 = 400 + (a_y)²$

$a_y = \sqrt{5225 } = 72.28 \ in$

$v_y (t) = 72.28t\\ y(t) = 72.28t²$

$x(4) = 5(4)³ - 105(4) = 320 - 420 = -100in\\ x(0) = 5(0)³ - 105(0) = 0in$

$y(4) =72.28(4)² = 1156.5 in\\y(0) = 0in$

$r(4) = \sqrt{(x)² +(y)²} = \\\sqrt{-100² + 1156.5²} = 1160.8\ in$

$r(0) = \sqrt{(x)² +(y)²} = \\\sqrt{0² + 0²} = 0\ in$

$\textsf{Total Velocity} = \dfrac{r(4) - r(0) }{ t(4) - t(0)} \\= \dfrac{1160.8}{4} = 290.2\ ins^{-2} = 19.8\ ft/s²$