Question

The displacement is given by x= t^3+2t^2+5 m, the acceleration (in m/s^2) at t=2 s is ?​

Answer 1

$\sf Given \begin {cases} & \sf { Displacement \: as \: a \: function \: of \: time \: = \: x. \: = \: t^3 + 2t^2 + 5t } \\ &\sf{Time \: = \: 2 \: sec } \\ \end {cases}$

⠀⠀⠀⠀

⠀⠀⠀⠀

$\sf To\: Find \begin {cases} & \sf { Acceleration \: at \: t_2 \: second \: =\: ??} \\ \end {cases}$

⠀⠀⠀⠀

⠀⠀⠀⠀

We know that :-

• Instantaneous velocity = Rate of change of displacement ⠀

⠀⠀⠀⠀⠀⠀⠀

$\star\;{\boxed{\sf{\pink{\dfrac{dx}{dt} = V }}}}$

⠀⠀⠀⠀

⠀⠀⠀⠀

$:\implies\sf x = t^3+ 2t^2 + 5$

⠀⠀⠀⠀

$:\implies\sf \dfrac{dx}{dt} = \dfrac{d\: t^3 + 2t^2 + 5 }{dt}$

⠀⠀⠀⠀

$:\implies\sf v = \dfrac{dt^3}{dt} + \dfrac{ d\: 2t^2}{dt}+ \dfrac{ d\: 5 }{dt}$

⠀⠀⠀⠀⠀⠀⠀

• Differentiation of a constant is zero .
• Differentiation of a logarithmic function is nx^(n-1) .

⠀⠀⠀⠀⠀⠀⠀

$:\implies\sf v = 3t^{3-1} + 2\times 2t^{2-1} + 0$

⠀⠀⠀⠀

$:\implies\sf v = 3t^2 + 4t$

⠀⠀⠀⠀

$\sf :\implies{\underline{\boxed{\bold{ v = 3t^2 \:+ \: 4t }}}}\;\bigstar$

⠀⠀⠀⠀

⠀⠀⠀⠀

• Instantaneous acceleration = Rate of change of velocity ⠀

⠀⠀⠀⠀⠀⠀

$\star\;{\boxed{\sf{\pink{\dfrac{dv}{dt} = a }}}}$

⠀⠀⠀⠀

⠀⠀⠀⠀

$:\implies\sf a = 3t^2+ 4t$

⠀⠀⠀⠀

$:\implies\sf \dfrac{dv}{dt} = \dfrac{d\: 3t^2 + 4t }{dt}$

⠀⠀⠀⠀

$:\implies\sf a = \dfrac{d\: 3t^2}{dt} + \dfrac{ d\: 4t}{dt}$

⠀⠀⠀⠀

⠀⠀⠀⠀⠀⠀⠀

$:\implies\sf a = 3\times 2 t^{2-1} + 4\times 1 \:t^{1-1}$

⠀⠀⠀⠀

$:\implies\sf a = 6t + 4$

⠀⠀⠀⠀

$\sf :\implies{\underline{\boxed{\bold{ a = 6t\: + 4 }}}}\;\bigstar$

⠀⠀⠀⠀

• Putting t = 2 ( given )

⠀⠀⠀⠀

$:\implies\sf a = 6\times 2 + 4$

⠀⠀⠀⠀

$:\implies\sf a = 12 + 4$

⠀⠀⠀⠀

$:\implies\sf a = 16$

⠀⠀⠀⠀

$\sf :\implies {\boxed{\red{\mathfrak{ Required \: acceleration \: = \: 16 m/s^2 }}}}$

⠀⠀⠀⠀

⠀⠀⠀_______________________________