Question

URGENT
A particle is moving along a straight line and its position is given by the relation x=(t^3 - 6t^2 - 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 the line.

Answer 1

Position is given by $x=t^3-6t^2-15t+40$

(a)
$velocity,\ v= \frac{dx}{dt}=3t^2-12t-15\\ v=0\\ \Rightarrow 3t^2-12t-15=0\\ \Rightarrow t^2-4t-5=0\\ \Rightarrow t^2-5t+t-5=0\\ \Rightarrow t(t-5)+1(t-5)=0\\ \Rightarrow (t+1)(t-5)=0\\ \Rightarrow t=5s\ \ \ \ \ \ \ \ (t \neq-1\ as\ t\ can't\ be\ negative)$

(b)
$at\ t=5s,\\x=5^3-6 \times 5^2-15 \times 5+40\\x=125-150-75+40\\x=-60m\\ \\at\ t=0,\\x=40m$

Position of particle at 5s = -60m
displacement = -60 - (40) = -100m.

(c)
$v=3t^2-12t-15\\ a= \frac{dv}{dt}=6t-12$

So acceleration is (6t-12) m/s²

Answer 2

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

$v = \frac{dx}{dt} = (3t^{2} - 12t + 15)m/sec$

$a = \frac{dv}{dt} = (6t - 12) m/sec^{2}$

$(a) \: 3t^{2} - 12t - 15 = 0\\\\Doing\: splitting\:middle\:term:\\\\\3t^{2} - 15t + 3t - 15 = 0$

$3(t - 5) + 3(t - 5) = 0\\\\So\:it\:can\:either\:be\:t=-1\:or\:t=5$

$As\:time\:cannot\:be\:negative\:t = -1\:is\:to\:be\:rejected.$

$Therefore,\:the\:time\:is\:5\:seconds.$

$(b)\:Position\:at\:t = 5\:seconds\:\:t = 0\:seconds\\\\x = 5^{3} - 6(5)^{2} - 15(5) + 40\\\\x = -\:60\:meters\\\\x = 40\:meters\\\\Displacement\:at\:t = 5\:seconds\:and\:t = 0\:seconds\\\\s = x5 - x0 \\\\Here, \:x5 = -60\:and\:x0 = 40\\\\s = - 60 - 40 \\\\s = - 100\:meters\\\\(c)\:Acceleration\:st\:time\:5\:seconds\\\\a = 6(5) - 12\\\\a = 30 - 12\\\\a =18\:m/sec^{2}$