Question

a particle moves along a straight line OX. At a time t[u = in seconds ] the distance x [in m] of the particle from O is given by x= 40+12t-t^3.how long would the particle travel before coming to rest?

Answer 1

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
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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= 12-3t²
at time t =t (the time when the particle comes to rest ):
v= 0;
=> 12-3t² = 0;
=> t = 2 s
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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
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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 travelled a distance of :
56-40 = 16 meters .
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hope this helps !

Answer 2

Answer:

The particle travels the distance before comes to rest will be equal to 16meters.

Explanation:

Consider that the particle has distance x from point O:

The given time -displacement equation:

               $x=40+12t-t^{3}$                         ..............(1)

Put t=0 in above equation

We get,    $x=40+ 12(0)-(0)^{3}\\ x=40m$

Consider after time 't' the velocity of particle will be zero.

Differentiate eq.(1) w.r.t. 't'

      $\frac{d(x)}{dt}=\frac{d}{dt} (40+12t-t^{3})\\ \\v=\frac{dx}{dt}=12-3t^{2}$                         ...............(2)

Put v=0 in eq. (2)

 ∴       $0=12-3t^{2}\\ 3t^{2}=12\\ t^{2}=4\\ t=2$

The distance covered by particle in  t=2s will be:

∴       $x=40+12(2)-(2)^{3}\\ x=40+24-8\\ x=56m$

Because particles starts moving when it was 40m away from point O

Then traveled distance before come to rest $=56-40=16m$