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Question

A particle moves along a straight line $$OX$$. At a time $$t$$ (in seconds) the distance $$x$$(in metres) of the particle from $$O$$ is given by $$x=40+12t-t^3$$. How long would the particle travel before coming to rest?


Solution

$$x = 40 +12t-t^3 $$
$$ \Rightarrow $$ velocity $$ u = \dfrac{dx}{dt} = \dfrac{d}{dt} (40+12t -t^3)$$
$$ \Rightarrow u = 0+12-3t^2 $$
rest $$ \Rightarrow  u =0 \Rightarrow  12-3t^2 = 0 \Rightarrow  t = 2 sec$$
So value of x at t = 2 sec 
$$ x = 40+12(2)-(2)^3$$
$$ \boxed{x = 56 \,meter.}$$

1187276_1103500_ans_efe3ee0eae304f9b9c2204df04ab9b9c.jpg

Physics

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