Question

# A particle moves along a straight line OX. At a time t(in seconds) the distance x(in meters)of the particle from 0 is given by:-$$x=40+12t-t^3$$.  How long would the particles travel before coming to rest?

A
16m
B
24m
C
40m
D
56m

Solution

## The correct option is A $$16 m$$At $$t=0 ,$$ particle is at $$,$$ let's say $$x$$ distance ,from $$O ;$$then putting $$t=0$$ in the given displacement-time equation we get$$;$$$$x =40 +12(0) -(0)^3 = 40 m$$Particle comes to rest that means velocity of particle becomes zero after travelling certain displacement $$;$$ let's say the time be $$t.$$then after differentiating the given displacement$$-$$time equation $$wrt.$$ time we get velocity$$-$$time equation$$v= 12-3t^2$$at time $$t =t$$ $$($$the time when the particle comes to rest $$):$$$$v= 0;$$$$=> 12-3t^2 = 0;$$$$=> t = 2 s$$Then $$,$$at $$t =2s$$ we are at $$,$$ let's say $$x'$$ distance from $$O ;$$put this value of$$t (=2)$$ in given displacement-time equation , we get$$;$$$$x'= 40 +12(2) -(2)^3;$$$$= 56m$$Further$$;$$We have seen that the particle started his journey when it is at $$40m$$ from the point $$O.$$ And came to rest at $$56m$$ from the point $$O .$$then the particle traveled a distance of$$:$$$$56-40=16m.$$Hence,option $$(A)$$ is correct answer.Physics

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