Question

Motion of a particle is given by equation s = (3t(cube) + 7t(sq.) + 14t + 8) m. the value of acceleration of the particle at t = 1 sec is

Answer 1

Given:

Position of particle w.r.t. time:

$\rm s = (3 {t}^{3} + 7 {t}^{2} + 14t + 8) \: m$

To Find:

Accelration (a) of particle at t = 1 s

Answer:

Double differentiation of position-time relation gives accelration:

$\bf \implies a = \dfrac{ {d}^{2}s }{d {t}^{2} } \\ \\ \rm \implies a = \dfrac{ {d}^{2} }{d {t}^{2} }(3 {t}^{3} + 7 {t}^{2} + 14t + 8) \\ \\ \rm \implies a = 18t + 14$

At t = 1 s:

$\rm \implies a = 18 \times 1 + 14 \\ \\ \rm \implies a = 18 + 14 \\ \\ \rm \implies a = 32 \: m {s}^{ - 2}$

$\therefore$ $\boxed{\mathfrak{Acceleration \ of \ particle \ at \ 1 \ s = 32 \ m/s^2}}$

Answer 2

$\;\;\underline{\textbf{\textsf{ Given:-}}}$

• $\rm s = (3 {t}^{3} + 7 {t}^{2} + 14t + 8) \: m$

$\;\;\underline{\textbf{\textsf{ To Find :-}}}$

• Acceleration, a when t = 1 sec

$\;\;\underline{\textbf{\textsf{ Solution :-}}}$

$\underline{\:\textsf{As we know the formula :}}$

$\bf \longrightarrow a = \dfrac{ {d}^{2}s }{d {t}^{2} }$

$\underline{\:\textsf{Substitute the value of s :}}$

$\\ \rm \longrightarrow a = \dfrac{ {d}^{2} }{d {t}^{2} }(3 {t}^{3} + 7 {t}^{2} + 14t + 8) \\ \\ \rm \longrightarrow a = 18t + 14$

$\underline{\:\textsf{Now, substitute t = 1 :}}$

$\rm \longrightarrow a = 18 \times 1 + 14 \\ \\ \rm \longrightarrow a = 18 + 14 \\ \\ \rm \longrightarrow a = 32 \: m {s}^{ - 2}$

$\;\;\underline{\textbf{\textsf{ Hence-}}}$

$\underline{\textsf{Acceleration of the particle at 1 sec is \textbf{32m/s²}}}.$

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