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.

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Answers

2015-07-20T19:30:00+05:30
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
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