Question

12. A particle is moving along a straight line and its position is given by the
relation
x=(t3 - 6t2- 15t +40) m
Find (a) The time at which velocity is zero.
(b) Position and displacement of the particle at that point.
(c) Acceleration for the particle at that line​

Answer 1

Position of the particle,

$\sf{\longrightarrow x=t^3-6t^2-15t+40}$

Initial position,

$\sf{\longrightarrow x(0)=0^3-6(0)^2-15(0)+40}$

$\sf{\longrightarrow x(0)=40}$

Then velocity,

$\sf{\longrightarrow v=\dfrac{dx}{dt}}$

$\sf{\longrightarrow v=\dfrac{d}{dt}(t^3-6t^2-15t+40)}$

$\sf{\longrightarrow v=3t^2-12t-15}$

For $\sf{v=0,}$

$\sf{\longrightarrow 3t^2-12t-15=0}$

$\sf{\longrightarrow t^2-4t-5=0}$

$\sf{\longrightarrow t^2+t-5t-5=0}$

$\sf{\longrightarrow t(t+1)-5(t+1)=0}$

$\sf{\longrightarrow (t+1)(t-5)=0}$

Since time is non - negative,

$\sf{\longrightarrow\underline{\underline{t=5\ s}}}$

Position of this particle at this time is,

$\sf{\longrightarrow x(5)=5^3-6(5)^2-15(5)+40}$

$\sf{\longrightarrow x(5)=125-6\times25-15\times5+40}$

$\sf{\longrightarrow\underline{\underline{x(5)=-60\ m}}}$

Displacement during this time will be,

$\sf{\longrightarrow s(5)=x(5)-x(0)}$

$\sf{\longrightarrow s(5)=-60-40}$

$\sf{\longrightarrow \underline{\underline{s(5)=-100\ m}}}$

Acceleration of the particle,

$\sf{\longrightarrow a=\dfrac{dv}{dt}}$

$\sf{\longrightarrow a=\dfrac{d}{dt}(3t^2-12t-15)}$

$\sf{\longrightarrow a=6t-12}$

At $\sf{t=5\ s,}$

$\sf{\longrightarrow a(5)=6(5)-12}$

$\sf{\longrightarrow\underline{\underline{a(5)=18\ m\,s^{-2}}}}$

Answer 2

Please refer to the above picture