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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
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B
24m
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C
40m
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D
56m
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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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