
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) 24m
B) 40m
C) 56m
D) 16m
Answer
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Hint: The distance is defined as the length of the path between the initial position and the final position of a moving object in the given interval of time. It is a scalar quantity. The change in the position of an object is called displacement. Displacement is a vector quantity and has both magnitude and direction.
Complete step by step solution:
Given equation \[ x = 40 + 12t - t^3\].....................(1)
Let say that the particle is at a distance x from O at a time, t=0
Thus substituting the value of t=0 in the given equation, \[x = 40 + 12t - {t^3}\], we get,
\[x = 40 + 12\left( 0 \right) - {\left( 0 \right)^3} = 40m\]
It is given that the particle will come to rest which means the velocity of the particle becomes zero after traveling to a certain displacement and let us consider this time as t.
Differentiating the given equation 1 to time, we get,
$\Rightarrow v = 12 - 3{t^2}$.................(2)
When the particle comes to rest at time=t, velocity=0
Substituting the value of v=0 in equation 2, we get,
$\Rightarrow 12 - 3{t^2} = 0$
$ \Rightarrow 3{t^2} = 12$
$ \Rightarrow {t^2} = 4$
$ \Rightarrow t = 2s$
Substituting the value of t in equation 1, we get
\[\Rightarrow {x} = 40 + 12\left( 2 \right) - {\left( 2 \right)^3}\]
\[ \Rightarrow {x} = 56m\]
Thus, it is clear that the particle had started the journey when it was at a distance of 40m from point O and came to rest at 56m from point O.
Thus, the particle had traveled a distance of $56m - 40m = 16m$.
Thus, the particle had traveled a distance of 16m before coming to rest.
Hence the correct option is D.
Note: 1. The velocity is defined as the rate of change of the position of the object to time. Velocity is also a vector quantity and has both direction and magnitude.
2. The distance and the displacement have the same SI unit, meter.
Complete step by step solution:
Given equation \[ x = 40 + 12t - t^3\].....................(1)
Let say that the particle is at a distance x from O at a time, t=0
Thus substituting the value of t=0 in the given equation, \[x = 40 + 12t - {t^3}\], we get,
\[x = 40 + 12\left( 0 \right) - {\left( 0 \right)^3} = 40m\]
It is given that the particle will come to rest which means the velocity of the particle becomes zero after traveling to a certain displacement and let us consider this time as t.
Differentiating the given equation 1 to time, we get,
$\Rightarrow v = 12 - 3{t^2}$.................(2)
When the particle comes to rest at time=t, velocity=0
Substituting the value of v=0 in equation 2, we get,
$\Rightarrow 12 - 3{t^2} = 0$
$ \Rightarrow 3{t^2} = 12$
$ \Rightarrow {t^2} = 4$
$ \Rightarrow t = 2s$
Substituting the value of t in equation 1, we get
\[\Rightarrow {x} = 40 + 12\left( 2 \right) - {\left( 2 \right)^3}\]
\[ \Rightarrow {x} = 56m\]
Thus, it is clear that the particle had started the journey when it was at a distance of 40m from point O and came to rest at 56m from point O.
Thus, the particle had traveled a distance of $56m - 40m = 16m$.
Thus, the particle had traveled a distance of 16m before coming to rest.
Hence the correct option is D.
Note: 1. The velocity is defined as the rate of change of the position of the object to time. Velocity is also a vector quantity and has both direction and magnitude.
2. The distance and the displacement have the same SI unit, meter.
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