Question

The motion of a particle is defined by the relation x = t4 - 10t2 + 8t + 12, where x and t are expressed in inches and seconds, respectively. Determine the position, the velocity, and the acceleration of the particle when t = 1s.

Answer 1

Answer:

2 inch/sec

- 8 inch/sec²

Explanation:

velocity = dx/dt

velocity = d(t^4 - 10t^2 + 8t + 12)/dt

velocity = 4t³ - 20t + 18

At = 1, velocity = 4(1)³ - 20(1) + 18 = 2

Velocity at t = 1 is 2 inch/sec

Acceleration = dv/dt

Acceleration = d(4t³ - 20t + 18)/dt

Acceleration = 12t² - 20

At t = 1, acceleration = 12(1)² - 20 = -8

Acceleration at t=1 is - 8inch/sec²

Answer 2

Stated that , The motion of a particle is defined by the relation x = t⁴ - 10t² + 8t + 12 where x and t are expressed in inches and seconds, respectively .

Need To Find : The Position , the Velocity , and Acceleration of the particle when t = 1 s ?

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━⠀

⠀⠀⠀⠀⠀⠀⠀⠀¤ Finding the Velocity of a particle :

$\qquad \dashrightarrow \sf x \:=\:t^4 \:- \:10t^2 \:+\:8t\:+\:12\:$

$\qquad \dashrightarrow \sf Velocity \:=\:\dfrac{dx}{dt}\:$

$\qquad \bigstar \:\:\underline {\pmb{ \purple {\sf By \: Differentiating \:\:the \:\: given \:Equation \:we \:get \:,\:}}}$

$\qquad \dashrightarrow \sf \: Velocity \:=\:\dfrac{dx}{dt}\:\\\\\\ \qquad \dashrightarrow \:\sf Velocity \:=\:\dfrac{d}{dt}\:\{ \:t^4 \:- \:10t^2 \:+\:8t\:+\:12\:\}\:\\\\\\ \qquad \dashrightarrow \sf \: Velocity \:=\: \:4t^3 \:- \:20t \:+\:\:18\:\:$

⠀⠀⠀⠀⠀▪︎⠀⠀The Velocity when t ( or time ) will be 1 second .

⠀⠀⠀⠀

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Value\:of \:t \::}}$

$\qquad \dashrightarrow \sf \: Velocity \:=\: \:4t^3 \:- \:20t \:+\:\:18\:\:\\\\\\ \qquad \dashrightarrow \:\sf Velocity \:=\: \:4(1)^3 \:- \:20(1) \:+\:\:18\:\:\\\\\\ \qquad \dashrightarrow \sf \: Velocity \:=\: \:4 \:- \:20 \:+\:\:18\:\:\\\\\\ \qquad \dashrightarrow \underline {\boxed {\pmb{\frak{ \: Velocity \:=\: \:2 \: inch\:/sec \:}}}}\:\:\bigstar \:\:$

⠀⠀⠀∴ Hence , the Velocity of Particle is 2 inch/sec .

⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━⠀

⠀⠀⠀⠀⠀⠀⠀⠀¤ Finding the Acceleration of a particle :

$\qquad \dashrightarrow \sf v \:=\:\:4t^3 \:- \:20t \:+\:\:18\:\:\:\:$

$\qquad \dashrightarrow \sf Acceleration \:=\:\dfrac{dv}{dt}\:$

$\qquad \bigstar\:\: \underline {\pmb{ \purple {\sf By \: Differentiating \:\:the \:\: \:Equation \:we \:get \:,\:}}}$

$\qquad \dashrightarrow \sf \: Acceleration \:=\:\dfrac{dv}{dt}\:\\\\\\ \qquad \dashrightarrow \:\sf Acceleration \:=\:\dfrac{d}{dt}\:\{ \:\:4t^3 \:- \:20t \:+\:\:18\:\:\:\}\:\\\\\\ \qquad \dashrightarrow \sf \: Acceleration \:=\: \: \:12t^2 \:-\:\:20\:\:$

⠀⠀⠀⠀⠀▪︎⠀⠀The Acceleration , when t ( or time ) will be 1 second .

⠀⠀⠀⠀

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Value\:of \:t \::}}$

$\qquad \dashrightarrow \sf \: Acceleration \:=\: \:12t^2 \:-\:\:20\:\:\\\\\\ \qquad \dashrightarrow \:\sf Acceleration \:=\: \:12(1)^2 \:-\:\:20\:\:\\\\\\ \qquad \dashrightarrow \sf \: Acceleration \:=\: \:12 \:-\:\:18\:\:\\\\\\ \qquad \dashrightarrow \underline {\boxed {\pmb{\frak{ \: Acceleration \:=\: \:-8 \: inch\:/sec^2 \:}}}}\:\:\bigstar \:\:$

⠀⠀⠀∴ Hence , the Acceleration of Particle is – 8 inch/sec² .