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 $$x = 40+12(2)-(2)^3$$$$\boxed{x = 56 \,meter.}$$Physics

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