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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.


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

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)

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.

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

So acceleration is (6t-12) m/s²
0 0 0
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